Limits and Continuity AP Calculus AB.

Limits and Continuity AP Calculus AB

Foerster: Exploration 2-1a
Instructional Focus: Evaluate a limit of a function graphically and numerically at a particle value of x and relate the limit to the definition.

Objective: Find the limit of a function that approaches an indeterminate form at a particular value of x and relate it to the definition.

Plot on your grapher the graph of this function.
𝑓 𝑥 = 𝑥 3 −7 𝑥 2 +17𝑥−15 𝑥−3 Use a friendly window with x = 3 as a grid point, but with the grid turned off. Sketch the results here. Show the behavior of the function in a neighborhood of x = 3. Chunking: Problem 1 This “friendly window” idea is tricky and varies from calculator to calculator. Zdecimal will work on the TI calculators for this problem and for windows that contain the origin. When setting a “friendly window” you only need to be concerned with the Xmin and Xmax. Emphasize there is a hole in the graph. You might want to have the students produce a table on their calculators.

Substitute 3 for x in the equation for f(x)
Substitute 3 for x in the equation for f(x). What form does the answer take? What name is given to an expression of this form? The graph of f has a removable discontinuity at x = 3. The y-value at this discontinuity is the limit of f(x) as x approaches 3. What number does this limit equal? Chunking: Problems 2-3

The result is called an indeterminate form.
Substitute 3 for x in the equation for f(x). What form does the answer take? What name is given to an expression of this form? 𝑓 3 = 27−63+51−15 3−3 = 0 0 The result is called an indeterminate form. The graph of f has a removable discontinuity at x = 3. The y-value at this discontinuity is the limit of f(x) as x approaches 3. What number does this limit equal? The limit is y = 2 as x approaches 3. It appears that the hole can be filled with the point (3, 2). Chunking: Problems 2-3

Make a table of values of f(x) for each 0.1 unit
change in x-value from 2.5 through 3.5. Between what two numbers does f(x) stay when x is kept in the open interval (2.5, 3.5)? Chunking: Problem 4-5 You might get the students to produce a table that “Zooms in” between 2.9 and 3.1 in order to see the f(x) approach the value 2. This question is preparing the way to the definition of limit at a point. If you had the students do a table between 2.9 and 3.1, ask them a similar question.

Simplify the fraction for f(x) and evaluate this simplified function for x = 3.
(x – 3) must be factor of x3 – 7x2 + 17x – 15 because when x = 3, both expressions have a value of zero. Factoring of x3 – 7x2 + 17x – 15 can be done through long division or other methods.

𝑓 𝑥 = 𝑥 3 −7 𝑥 2 +17𝑥−15 𝑥−3 = 𝑥−3 𝑥 2 −4𝑥+5 𝑥−3 = 𝑥 2 −4𝑥+5
Simplify the fraction for f(x) and evaluate this simplified function for x = 3. 𝑓 𝑥 = 𝑥 3 −7 𝑥 2 +17𝑥−15 𝑥−3 = 𝑥−3 𝑥 2 −4𝑥+5 𝑥−3 = 𝑥 2 −4𝑥+5 𝑓 3 = 3 2 −4 3 +5=9−12+5=2 (x – 3) must be factor of x3 – 7x2 + 17x – 15 because when x = 3, both expressions have a value of zero. Factoring of x3 – 7x2 + 17x – 15 can be done through long division or other methods.

2.2 – Limits: A Numerical and Graphical Approach

The goal in this section is to define limits and study them using numerical and graphical techniques. We begin with the following question: How do the values of a function f(x) behave when x approaches a number c, whether or not f(c) is defined?

To explore this question, we’ll experiment with the function
𝑓 𝑥 = sin 𝑥 𝑥 Evaluate f(0) Be sure x is in radians.

To explore this question, we’ll experiment with the function
𝑓 𝑥 = sin 𝑥 𝑥 𝑓 0 = sin 0 0 = 0 0 This is called an indeterminate form We can compute f(x) for values of x close to 0. Be sure x is in radians.

The notation for this is 𝑥→0, and more specifically we write
To describe the trend, we use the phrase “x approaches 0” to indicate that x takes on values (both positive and negative) that get closer and closer to 0. The notation for this is 𝑥→0, and more specifically we write 𝑥→ 0 + if x approaches 0 from the right 𝑥→ 0 − if x approaches 0 from the left

Complete the following table to determine the value f(x) approaches as x approaches 0.
sin 𝒙 𝒙 1 -1 0.5 -0.5 0.1 -0.1 0.05 -0.05 0.01 -0.01 0.005 -0.005 0.001 -0.001 𝑥→ 0 + 𝑓 𝑥 →____ 𝑥→ 0 −

Complete the following table to determine the value f(x) approaches as x approaches 0.
sin 𝒙 𝒙 1 -1 0.5 -0.5 0.1 -0.1 0.05 -0.05 0.01 -0.01 0.005 -0.005 0.001 -0.001 𝑥→ 0 + 𝑓 𝑥 →1 𝑥→ 0 − The table gives the unmistakable impression that f(x) gets closer and closer to 1 as 𝑥→ 0 + and as 𝑥→ 0 − .

The conclusion found from the table of values is supported by the graph of f(x).
The point (0, 1) is missing from the graph because f(x) is not defined at x = 0 However, the graph approaches this missing point as x approaches 0 from the left and right. We say that the limit of f(x) as x ⟶ 0 is equal to 1 lim 𝑥→0 𝑓 𝑥 =1

Investigate the following limit by completing the table
𝒙→ 𝟗 − 𝒙−𝟗 𝒙 −𝟑 𝒙→ 𝟗 + 8.9 9.1 8.99 9.01 8.999 9.001 8.9999 9.0001

Caution: Numerical investigations are often suggestive, but may be misleading in some cases.
Using tables and looking at graphs can give us a good idea of what is happening and what it appears as though the limit is approaching. However, they should not be used to determine an exact limit In this unit we will discover methods to determine exact limits.

The limits discussed so far are two=sided
To show that lim 𝑥→𝑐 𝑓 𝑥 =𝐿 , we should check that f(x) converges to L as x approaches c through values both larger and smaller than c That is, we must show lim 𝑥→ 𝑓 𝑥 = lim 𝑥→ 0 − 𝑓 𝑥 =𝐿 In some instances, f(x) may approach L from one side of c without necessarily approaching it from the other side, or f(x) may be defined on only ones side of c. The limit itself exists if both one-sided limits exist and are equal.

Investigate the one-sided limits of 𝑓 𝑥 = 𝑥 𝑥 as 𝑥→0.
Does lim 𝑥→0 𝑓 𝑥 exist?

The function f(x) in the figure below is not defined at c = 0, 2, 4
The function f(x) in the figure below is not defined at c = 0, 2, 4. Investigate the one- and two-sided limits at these points.

Some functions f(x) tend to ∞ or -∞ as x approaches a value c
In these cases, lim 𝑥→𝑐 𝑓 𝑥 does not exist, but we say that f(x) has an infinite limit. lim 𝑥→𝑐 𝑓 𝑥 =∞ if f(x) increases without bound as x ⟶ c lim 𝑥→𝑐 𝑓 𝑥 =−∞ if f(x) decreases without bound as x ⟶ c Keep in mind that ∞ and -∞ are not numbers and the above notation means that the limit does not exist because the graph increases (or decreases) without bound When f(x) approaches ∞ or -∞ as x approaches c from one or both sides, the line x = c is called a vertical asymptote.

Investigate the limits graphically:
lim 𝑥→ ln 𝑥 lim 𝑥→0 sin 𝜋 𝑥

lim 𝑥→ ln 𝑥

lim 𝑥→0 sin 𝜋 𝑥

If f(x) approaches a limit as x ⟶ c, then the limit value L is unique.
Summary If f(x) approaches a limit as x ⟶ c, then the limit value L is unique. A limit does not exist if: lim 𝑥→ 𝑐 − 𝑓 𝑥 ≠ lim 𝑥→ 𝑐 + 𝑓 𝑥 : The one-sided limits are not equal lim 𝑥→𝑐 𝑓 𝑥 =±∞ : The graph increases or decreases without bound f(x) oscillates The limit may exist even if f(c) is not defined. If a one- or two-sided limit is ±∞, the vertical line x = c is called a vertical asymptote.

Warm-Up! Evaluate the limit lim 𝑥→3 𝑥 2 −𝑥−6 2 𝑥 2 −3𝑥−9

2.3 – Basic Limit Laws

In the previous section we relied on graphical and numerical approaches to investigate limits and estimate their values. In the next sections we go beyond this intuitive approach and develop tools for computing limits in a precise way.

Basic Limit Laws If lim 𝑥→𝑐 𝑓 𝑥 and lim 𝑥→𝑐 𝑔 𝑥 exist, then
Sum and Difference Law: lim 𝑥→𝑐 𝑓 𝑥 ±𝑔 𝑥 exists and lim 𝑥→𝑐 𝑓 𝑥 ±𝑔 𝑥 = lim 𝑥→𝑐 𝑓 𝑥 ± lim 𝑥→𝑐 𝑔 𝑥 Constant Multiple Law: For any number k, lim 𝑥→𝑐 𝑘𝑓 𝑥 exists and lim 𝑥→𝑐 𝑘𝑓 𝑥 =𝑘 lim 𝑥→𝑐 𝑓 𝑥 Product Law: lim 𝑥→𝑐 𝑓 𝑥 𝑔 𝑥 exists and lim 𝑥→𝑐 𝑓 𝑥 𝑔 𝑥 = lim 𝑥→𝑐 𝑓 𝑥 lim 𝑥→𝑐 𝑔 𝑥

Basic Limit Laws Continued
If lim 𝑥→𝑐 𝑓 𝑥 and lim 𝑥→𝑐 𝑔 𝑥 exist, then Quotient Law: If lim 𝑥→𝑐 𝑔 𝑥 ≠0, then lim 𝑥→𝑐 𝑓 𝑥 𝑔 𝑥 exists and lim 𝑥→𝑐 𝑓 𝑥 𝑔 𝑥 = lim 𝑥→𝑐 𝑓 𝑥 lim 𝑥→𝑐 𝑔 𝑥 Powers and Roots: If p, q are integers with q ≠ 0, then lim 𝑥→𝑐 𝑓 𝑥 𝑝/𝑞 exists and lim 𝑥→𝑐 𝑓 𝑥 𝑝/𝑞 = lim 𝑥→𝑐 𝑓 𝑥 𝑝/𝑞

For any constants k and c,
Theorem For any constants k and c, lim 𝑥→𝑐 𝑘 =𝑘 lim 𝑥→𝑐 𝑥 =𝑐

Some limits can be evaluated by a simple substitution.
Substitution is valid when the function is continuous, a concept we shall study later. Try direct substitution as the first step in evaluating limits.

Examples lim 𝑥→2 𝑥 3 lim 𝑥→2 𝑥 3 +5𝑥+7 lim 𝑥→2 𝑥 3 +5𝑥+7 lim 𝑡→−1 𝑡+6 2 𝑡 4 lim 𝑡→3 𝑡 −1/4 𝑡+5 1/3

Examples lim 𝑥→2 𝑥 3 = 2 3 =8 lim 𝑥→2 𝑥 3 +5𝑥+7 = =8+10+7=25 lim 𝑥→2 𝑥 3 +5𝑥+7 = = = 25 =5 lim 𝑡→−1 𝑡+6 2 𝑡 4 = − −1 4 = 5 2 =2.5 lim 𝑡→3 𝑡 −1/4 𝑡+5 1/3 = 3 −1/ /3 = / /3 =

Warm-up Sketch the graph of the function. 𝑓 𝑥 = 2𝑥+1 𝑥<1 𝑥−1 𝑥≥1
𝑓 𝑥 = 2𝑥+1 𝑥<1 𝑥−1 𝑥≥1 𝑓 𝑥 = 𝑥 𝑥≤2 2𝑥 𝑥>2 Which graph appears to be continuous?

2.4 – Limits and Continuity

On a white board, take one minute to write down what you think a formal definition for a function to be continuous could be. Include what it would mean for a function to be discontinuous. Take one minute to share with a partner. Each pair will take one minute to share with the class.

In everyday speech, the word “continuous” means having no breaks or interruptions.
In calculus, continuity is used to describe functions whose graphs have no breaks.

Look at graph A below. Determine f(2) and lim 𝑥→2 𝑓 𝑥 .

Look at graph B below. Determine f(2) and lim 𝑥→2 𝑓 𝑥 .

Look at graph C below. Determine f(2) and lim 𝑥→2 𝑓 𝑥 .

Look at graph D below. Determine f(2) and lim 𝑥→2 𝑓 𝑥 .

Look at graph E below. Determine f(2) and lim 𝑥→2 𝑓 𝑥 .

Look at graph F below. Determine f(2) and lim 𝑥→2 𝑓 𝑥 .

Complete the table from the information found.
Graph f(2) lim 𝒙→𝟐 𝒇 𝒙 Continuous? A B C D E F

Complete the table from the information found.
Graph f(2) lim 𝒙→𝟐 𝒇 𝒙 Continuous? A 6 DNE No B 4 C Yes D E F

There are three conditions that must hold for a function to be continuous. Based on the table above, what do you think those three conditions are?

Definition – Continuity at a Point
Assume that f(x) is defined on an open interval containing x = c. Then f is continuous at x = c if lim 𝑥→𝑐 𝑓 𝑥 =𝑓 𝑐 If the limit does not exist, or if it exists but is not equal to f(c), we say that f has a discontinuity (or is discontinuous) at x = c.

A function f(x) may be continuous at some points and discontinuous at others.
If f(x) is continuous at all points in an interval I, then f(x) is said to be continuous on I. If f(x) is continuous at all points in its domain, then f(x) is simply called continuous.

Keep in mind that continuity at a point x = c requires more than just the existence of a limit.
Three conditions must hold: f(c) is defined lim 𝑥→𝑐 𝑓 𝑥 exists They are equal If lim 𝑥→𝑐 𝑓 𝑥 exists but is not equal to f(c), we say that f has a removable discontinuity at x = c. We say that f(x) has a nonremovable discontinuity at x = c if one or both of the one-sided limits is infinite.

Theorem – Continuity of the Inverse Function
If f(x) is continuous on an interval I with range R, and if 𝑓 −1 𝑥 exists, then 𝑓 −1 𝑥 is continuous with domain R. Theorem – Continuity of Composite Functions If g is continuous at x = c, and f is continuous at x = g(c), then the composite function F(x) = f(g(x)) is continuous at x = c.

Definition: f(x) is continuous at x = c if lim 𝑥→𝑐 𝑓 𝑥 =𝑓 𝑐
Summary Definition: f(x) is continuous at x = c if lim 𝑥→𝑐 𝑓 𝑥 =𝑓 𝑐 If lim 𝑥→𝑐 𝑓 𝑥 does not exist, or if it exists but does not equal f(c), then f is discontinuous at x = c If f(x) is continuous at all points in its domain, f is simply called continuous Three common types of discontinuities: Removable discontinuity: the limit exists but does not equal f(c) Jump discontinuity: the one-sided limits both exist but are not equal Infinite discontinuity: the limit is infinite as x approaches c

2.5 – Evaluating Limits Algebraically

Substitution can be used to evaluate limits when the function in question is known to be continuous.
When we study derivatives, we will be faced with limits where f(c) is not defined. Many of these limits can be evaluated if we use algebra to rewrite the formula for f(x). Our strategy when direct substitution gives us an indeterminate form, is to transform f(x) algebraically into a new expression that is defined and continuous at x = c, and then evaluate the limit by substitution.

Examples lim 𝑥→3 𝑥 2 −4𝑥+3 𝑥 2 +𝑥−12 lim 𝑥→3 𝑥 2 −4𝑥+3 𝑥 2 +𝑥−12 = lim 𝑥→3 𝑥−3 𝑥−1 𝑥+4 𝑥−3 = lim 𝑥→3 𝑥−1 𝑥+4 = 3−1 3+4 = 2 7 lim 𝑥→2 𝑥 2 −4 𝑥 2 −6𝑥+8 lim 𝑥→2 𝑥 2 −4 𝑥 2 −6𝑥+8 = lim 𝑥→2 𝑥+2 𝑥−2 𝑥−2 𝑥−4 = lim 𝑥→2 𝑥+2 𝑥−4 = 2+2 2−4 = 4 −2 =−2 lim 𝑥→ 𝜋 tan 𝑥 sec 𝑥 lim 𝑥→ 𝜋 tan 𝑥 sec 𝑥 = lim 𝑥→ 𝜋 sin 𝑥 cos 𝑥 cos 𝑥 = lim 𝑥→ 𝜋 sin 𝑥 cos 𝑥 ∙ cos 𝑥 1 = lim 𝑥→ 𝜋 2 sin 𝑥 = sin 𝜋 2 =1

Theorem – Important Trigonometric Limits
lim 𝜃→0 sin 𝜃 𝜃 = lim 𝜃→0 1− cos 𝜃 𝜃 =0

Examples lim 𝜃→0 2 sin 𝜃 cos 𝜃 𝜃 lim 𝜃→0 2 sin 𝜃 cos 𝜃 𝜃 = lim 𝜃→0 2∙ sin 𝜃 𝜃 ∙ cos 𝜃 = lim 𝜃→ lim 𝜃→0 sin 𝜃 𝜃 lim 𝜃→0 cos 𝜃 =2∙1∙1=2 lim 𝑥→0 sin 𝑥 − sin 𝑥 cos 𝑥 𝑥 2 lim 𝑥→0 sin 𝑥 − sin 𝑥 cos 𝑥 𝑥 2 = lim 𝑥→0 sin 𝑥 1− cos 𝑥 𝑥 2 = lim 𝑥→0 sin 𝑥 𝑥 ∙ 1− cos 𝑥 𝑥 = lim 𝑥→0 sin 𝑥 𝑥 lim 𝑥→0 1− cos 𝑥 𝑥 =1∙0=0

Limits Review

Complete the table and use the result to estimate the limit.
x 1.9 1.99 1.999 2.001 2.01 2.1 f(x) x -0.1 -0.01 -0.001 0.001 0.01 0.1 f(x)

Complete the table and use the result to estimate the limit.
x 1.9 1.99 1.999 2.001 2.01 2.1 f(x) .34483 .33445 .33344 .33322 .33223 .32258 x -0.1 -0.01 -0.001 0.001 0.01 0.1 f(x) .29112 .28892 .28869 .28865 .28843 .28631

Use the graph to find the limit (if it exists)
Use the graph to find the limit (if it exists). If the limit does not exist, explain why. lim 𝑥→2 5−𝑥

Use the graph to find the limit (if it exists). If the limit does not exist, explain why. lim 𝑥→2 5−𝑥 =3

Use the graph to find the limit (if it exists). If the limit does not exist, explain why. For 𝑓 𝑥 = 3−𝑥 𝑥≠ 𝑥=1 , find lim 𝑥→1 𝑓 𝑥

Use the graph to find the limit (if it exists). If the limit does not exist, explain why. For 𝑓 𝑥 = 3−𝑥 𝑥≠ 𝑥=1 , find lim 𝑥→1 𝑓 𝑥 =2

State the three types of behavior associated with nonexistence of a limit and draw a function to represent each.

State the three types of behavior associated with nonexistence of a limit and draw a function to represent each. lim 𝑥→ 𝑐 − 𝑓 𝑥 ≠ lim 𝑥→ 𝑐 + 𝑓 𝑥 : The one-sided limits do not approach the same number. lim 𝑥→𝑐 𝑓 𝑥 =±∞: The graph increases or decreases without bound. The graph oscillates as 𝑥→𝑐

Sketch a graph of a function f that satisfies the given values.
𝑓 0 is undefined lim 𝑥→0 𝑓 𝑥 =4 𝑓 2 =6 lim 𝑥→2 𝑓 𝑥 =3

Sketch a graph of a function f that satisfies the given values.
𝑓 0 is undefined – This says there is a hole when x = 0 lim 𝑥→0 𝑓 𝑥 =4 – This clarifies that the hole is at (0, 4) and that the graph should approach 4 from both sides. 𝑓 2 =6 – This says there is a point at (2, 6) lim 𝑥→2 𝑓 𝑥 =3 – This says there is a hole at (2, 3) but that the graph should approach 3 from both sides.

Find the limit, if it exists.
2∙0−1=−1 lim 𝑥→1 𝑥−3 𝑥 2 +4 1− = −2 5 lim 𝑥→ 𝜋 2 sin 𝑥 sin 𝜋 2 =1

Find the limit, if it exists. lim 𝑥→0 sec 2𝑥
sec 2∙0 = sec 0 = 1 cos 0 = 1 1 =1 lim 𝑥→3 tan 𝜋𝑥 4 tan 3𝜋 4 =−1 lim 𝑥→5 𝑥−5 𝑥 2 −25 5− −25 = 0 0 indeterminate form; try a different method; lim 𝑥→5 𝑥−5 𝑥+5 𝑥−5 = lim 𝑥→5 1 𝑥+5 = = 1 10

Find the limit, if it exists lim 𝑥→−1 𝑥 2 −1 𝑥+1
−1 2 −1 −1+1 = 0 0 indeterminate form; try a different method; lim 𝑥→−1 𝑥+1 𝑥−1 𝑥+1 = lim 𝑥→−1 𝑥−1 =−1−1=−2 lim 𝑥→−6 −𝑥−6 𝑥 2 −36 6−6 −6 2 −36 = 0 0 indeterminate form; try a different method; lim 𝑥→−6 −1 𝑥+6 𝑥+6 𝑥−6 = lim 𝑥→−6 −1 𝑥−6 = −1 −6−6 = 1 12

Find the limit, if it exists. lim 𝑥→0 sin 𝑥 1− cos 𝑥 2 𝑥 2
sin 0 1− cos 0 2∙ 0 2 = 0 1−1 0 = 0 0 indeterminate form; try a different method; lim 𝑥→0 sin 𝑥 𝑥 1− cos 𝑥 𝑥 = =0 lim 𝑥→0 cos 𝑥 tan 𝑥 𝑥 lim 𝑥→0 cos 𝑥 sin 𝑥 cos 𝑥 𝑥 = lim 𝑥→0 sin 𝑥 𝑥 =1

Graph the function and determine if the function is continuous, using correct math notation, at x = 1. 𝑓 𝑥 = 𝑥 𝑥>1 𝑥+4 𝑥≤1 lim 𝑥→ 𝑓 𝑥 = lim 𝑥→ 𝑥 = =5 lim 𝑥→ 1 − 𝑓 𝑥 = lim 𝑥→ 1 − 𝑥+4 =1+4=5 Because lim 𝑥→ 𝑓 𝑥 =5= lim 𝑥→ 1 − 𝑓 𝑥 , lim 𝑥→1 𝑓 𝑥 =5 𝑓 1 =1+4=5 Since lim 𝑥→1 𝑓 𝑥 =5=𝑓 1 , f(x) is continuous at x = 1.

Graph the function and determine if the function is continuous, using correct math notation, at x = 1. 𝑓 𝑥 = 𝑥 2 −1 𝑥≥1 2𝑥−1 𝑥<1 lim 𝑥→ 𝑓 𝑥 = lim 𝑥→ 𝑥 2 −1 = 1 2 −1=0 lim 𝑥→ 1 − 𝑓 𝑥 = lim 𝑥→ 1 − 2𝑥−1 =2 1 −1=1 Since lim 𝑥→ 𝑓 𝑥 ≠ lim 𝑥→ 1 − 𝑓 𝑥 , lim 𝑥→1 𝑓 𝑥 does not exist and therefore f(x) is not continuous at x = 1.

Find the value of c that makes the function continuous.
𝑓 𝑥 = 𝑥 2 +𝑐 𝑥>2 𝑥−𝑐 𝑥≤2 2 2 +𝑐=2−𝑐 4+𝑐=2−𝑐 2𝑐=−2 𝑐=−1 𝑓 𝑥 = 𝑥 2 −1 𝑥>2 𝑥 𝑥≤2

Find the value of c that makes the function continuous.
𝑓 𝑥 = 2𝑥+3𝑐 𝑥≥4 𝑥 +𝑐 𝑥<4 2∙4+3𝑐= 4 +𝑐 8+3𝑐=2+𝑐 2𝑐=−6 𝑐=−3 𝑓 𝑥 = 2𝑥−9 𝑥≥4 𝑥 −3 𝑥<4

Limits and Continuity Quiz

Exploration 2-4a: Continuous and Discontinuous Functions

Let f be the piecewise function defined by
𝑓 𝑥 = 𝑥+1, if 𝑥<2 𝑘 𝑥−5 2 , if 𝑥≥2 where k stands for a constant. Plot the graph of f for k = 1. Sketch the result. Function f is discontinuous at x = 2. Tell what it means for a function to be discontinuous. Chunking: Problems 1-2

𝑓 𝑥 = 𝑥+1, if 𝑥<2 𝑘 𝑥−5 2 , if 𝑥≥2 where k stands for a constant. Plot the graph of f for k = 1. Sketch the result. Chunking: Problems 1-2

𝑓 𝑥 = 𝑥+1, if 𝑥<2 𝑘 𝑥−5 2 , if 𝑥≥2 where k stands for a constant. Function f is discontinuous at x = 2. Tell what it means for a function to be discontinuous. f(a) does not exist, or lim 𝑥→𝑎 𝑓 𝑥 does not exist, or lim 𝑥→𝑎 𝑓 𝑥 ≠𝑓 𝑎 Chunking: Problems 1-2

Find lim 𝑥→ 2 − 𝑓 𝑥 and lim 𝑥→ 2 + 𝑓 𝑥
Find lim 𝑥→ 2 − 𝑓 𝑥 and lim 𝑥→ 𝑓 𝑥 . (The second limit will be in terms of k.) what must be true of these two limits for f to be continuous at x = 2? Find the value of k that makes f continuous at x = 2. Sketch the graph of f for this value of k. Chunking: Problems 3-4

lim 𝑥→ 2 − 𝑓 𝑥 = lim 𝑥→ 2 − 𝑥+1 =2+1=3
Find lim 𝑥→ 2 − 𝑓 𝑥 and lim 𝑥→ 𝑓 𝑥 . (The second limit will be in terms of k.) what must be true of these two limits for f to be continuous at x = 2? lim 𝑥→ 2 − 𝑓 𝑥 = lim 𝑥→ 2 − 𝑥+1 =2+1=3 lim 𝑥→ 𝑓 𝑥 = lim 𝑥→ 𝑘 𝑥−5 2 =𝑘 2−5 2 =9𝑘 Find the value of k that makes f continuous at x = 2. Sketch the graph of f for this value of k. 3=9𝑘 𝑘= 1 3 Chunking: Problems 3-4

The graph in Problem 4 has a cusp at x = 2
The graph in Problem 4 has a cusp at x = 2. What is the origin of the word cusp, and why is it appropriate to use in this context? Suppose someone asks, “Is f(x) increasing or decreasing at x = 2 with k as in Problem 4?” How would you have to answer that question? What, then, can you conclude about the derivative of a function at a point where the graph has a cusp? Chunking: Problems 5-6

The graph in Problem 4 has a cusp at x = 2
The graph in Problem 4 has a cusp at x = 2. What is the origin of the word cusp, and why is it appropriate to use in this context? Suppose someone asks, “Is f(x) increasing or decreasing at x = 2 with k as in Problem 4?” How would you have to answer that question? What, then, can you conclude about the derivative of a function at a point where the graph has a cusp? Chunking: Problems 5-6 A cusp is a point on the graph at which the function is continuous but the derivative is not continuous. A cusp is a sharp point or an abrupt change in direction. Allow students sufficient time to discuss problem 6 with each other and form an answer. Then have several share. Lead a discussion of differentiability at a point. A function is not differentiable at a point if: Not continuous Has a cusp, sharp corner Or has a vertical tangent line

Differentiability

Defintion: A function f is differentiable at a if 𝑓 ′ 𝑎 exists
𝑓 ′ 𝑎 = lim 𝑥→𝑎 𝑓 𝑥 −𝑓 𝑎 𝑥−𝑎 So, how can a function fail to be differentiable? When the limit does not exist. So, how can a limit fail to exist? The right and left hand limits do not agree or the limit is infinity

So, the slope 𝑓 𝑥 −𝑓 𝑎 𝑥−𝑎 to the right and left do not agree.
This is called a cusp. It is like a corner or kink in the graph. Or the slope is infinite or there is a vertical tangent line.

Differentiability is connected to local linearity.
If f is differentiable at x = a, then when we zoom in at the point (a, f(a)), the graph straightens out and appears more and more like a line.

At a cusp, no amount of zooming will straighten out the graph.
At a discontinuity, no amount of zooming will straighten out the graph even though the left and right rates of change may be the same.

Inverse: If not p, then not q. Contrapositive: If not q, then not p.
Refresher Statement: If p, then q. Converse: If q, then p. Inverse: If not p, then not q. Contrapositive: If not q, then not p. If the statement is true, the contrapositive must be true. If the converse is true, the inverse must be true. Use p as it is snowing and q as they cancel school.

Differentiability Implies Continuity
If f is differentiable at x = a, then f is continuous at x = a. What is the contrapositive to this statement? If f is not continuous at x = a, then f is not differentiable at x = a. Both of these statements are true.

Differentiability Implies Continuity
If f is differentiable at x = a, then f is continuous at x = a. What is the converse statement? Is it true? Why or why not? If f is continuous at x = a, then f is differentiable at x = a. What is the inverse statement? Is it true? Why or why not? If f is not differentiable at x = a, then f is not continuous at x = a.

But, if a function f is continuous at x = a, then f is not necessarily differentiable at x = a
Or, if a function f is not differentiable at x = a, then f may be continuous at x = a. Therefore, the converse and inverse statements are not true.

Differentiability and Continuity Practice

State whether the graph illustrates a function that
has left and right limits at the marked value of x. has a limit at the marked value of x. is continuous at the marked value of x. If it is not continuous there, explain why. is differentiable at the marked value of x. If it is not differentiable there, explain why. The function has right and left limits but the limit does not exist because the two limits are not equal. Therefore the function is not continuous. The function is not differentiable because the derivative from the right side is undefined.

has left and right limits at the marked value of x. has a limit at the marked value of x. is continuous at the marked value of x. If it is not continuous there, explain why. is differentiable at the marked value of x. If it is not differentiable there, explain why. The function has a limit because it has right and left side limits that are equal. The function is not continuous because f(3) is not defined. Since f(3) is not defined, f’(3) cannot exist.

has left and right limits at the marked value of x. has a limit at the marked value of x. is continuous at the marked value of x. If it is not continuous there, explain why. is differentiable at the marked value of x. If it is not differentiable there, explain why. All limits exist and the function is continuous. The function is not differentiable because there is a sharp point (cusp) at x = 4. f’(4) is not defined because the derivative from the left is positive and the derivative from the right is negative.

has left and right limits at the marked value of x. has a limit at the marked value of x. is continuous at the marked value of x. If it is not continuous there, explain why. is differentiable at the marked value of x. If it is not differentiable there, explain why. The function is continuous and differentiable at x = -3.

has left and right limits at the marked value of x. has a limit at the marked value of x. is continuous at the marked value of x. If it is not continuous there, explain why. is differentiable at the marked value of x. If it is not differentiable there, explain why. Neither right or left limits exist because there is a vertical asymptote at x = 2. The function is not continuous and undefined at x = 2, so it is not differentiable.

has left and right limits at the marked value of x. has a limit at the marked value of x. is continuous at the marked value of x. If it is not continuous there, explain why. is differentiable at the marked value of x. If it is not differentiable there, explain why. The function is continuous at differentiable at x = 5.

has left and right limits at the marked value of x. has a limit at the marked value of x. is continuous at the marked value of x. If it is not continuous there, explain why. is differentiable at the marked value of x. If it is not differentiable there, explain why. The function has a limit because it has right and left side limits that are equal. The function is not continuous because f(1) is not the same value as the limit. The function is not differentiable at x = 1.

has left and right limits at the marked value of x. has a limit at the marked value of x. is continuous at the marked value of x. If it is not continuous there, explain why. is differentiable at the marked value of x. If it is not differentiable there, explain why. The function has right and left limits but the limit does not exist because the two limits are not equal. Therefore the function is not continuous. The function is not differentiable because the derivative from the left side is undefined.

For the graphs below, list the x-values for which the function appears to be (i) not continuous and (ii) not differentiable. Give a reason for your answer. Not continuous at x = 1 because the limit does not exist. Also not differentiable at x = 1 because it is not continuous. The function is also not differentiable at x = 2 and x = 3 because there is a cusp. At a cusp the right and left sided derivatives are not equal.

For the graphs below, list the x-values for which the function appears to be (i) not continuous and (ii) not differentiable. Give a reason for your answer. The function is continuous at all points. The function is not differentiable at x = 2 because the tangent line appears to be vertical. The function is also not differentiable at x = 4 because there is a cusp.

For the functions graphed below, does the function appear to be differentiable on the interval of x-values shown. The function does not appear to be differentiable on the interval of x-values shown because there are two sharp corners (cusps).

For the functions graphed below, does the function appear to be differentiable on the interval of x-values shown. This function appears to be differentiable on the interval of x-values shown.

Explain what is wrong with each of the following statements.
A function f that is not differentiable at x = 0 has a graph with a sharp corner at x = 0. Besides a sharp corner, the function could be not continuous or have a vertical tangent line.

Explain what is wrong with each of the following statements.
If f is not differentiable at a point then it is not continuous at that point. A function can be continuous but just have a sharp corner at a point, like f(x) = |x| at x = 0.

Give an example graphically or algebraically of a continuous function that is not differentiable at x = 2. Explain your reasoning. One example: f(x) = |x – 2|

Give an example graphically or algebraically of a function not continuous at x = 2 that is differentiable at x = 2. Explain your reasoning. Not possible since if a function is not continuous it cannot be differentiable.

Are the statements below true or false
Are the statements below true or false? If a statement is true, give an example illustrating it. If a statement is false, give a counterexample. There is a function which is continuous on [1, 5] but not differentiable at x = 3. True. Sample answer: f(x) = |x – 3|

Are the statements below true or false? If a statement is true, give an example illustrating it. If a statement is false, give a counterexample. If a function is differentiable, then it is continuous. True.

Are the statements below true or false? If a statement is true, give an example illustrating it. If a statement is false, give a counterexample. If a function is continuous, then it is differentiable. False. Sample answer: f(x) = |x – 3|

Are the statements below true or false? If a statement is true, give an example illustrating it. If a statement is false, give a counterexample. If a function is not continuous, then it is not differentiable. True

Are the statements below true or false? If a statement is true, give an example illustrating it. If a statement is false, give a counterexample. If a function is not differentiable then it is not continuous. False. Sample answer: f(x) = |x – 3|

Exploration: Intermediate Value Theorem

The figure shows the graph of a function f defined on the closed interval [1, 5]. Why does the function f not meet the hypotheses of the intermediate value theorem? For function f in problem 1, explain why the conclusion of the intermediate value theorem is false if y = 5? For function f in problem 1, is there a value of x for which y = 3? Does this fact contradict the intermediate value theorem? Chunking: Problems 1-3 The purpose of these three problems is to show that the Intermediate Value Theorem conclusion might be true but the conclusion is only guaranteed if the function is continuous. In problems 1-3, the function f is not continuous and neither y= 5 not y = 3 are guaranteed but there is a number c where f(c) = 3.

The figure shows the graph of a function f defined on the closed interval [1, 5]. Why does the function f not meet the hypotheses of the intermediate value theorem? The hypotheses states that f must be continuous and at x = 3, the function f has a jump discontinuity. Chunking: Problems 1-3 The purpose of these three problems is to show that the Intermediate Value Theorem conclusion might be true but the conclusion is only guaranteed if the function is continuous. In problems 1-3, the function f is not continuous and neither y= 5 not y = 3 are guaranteed but there is a number c where f(c) = 3.

For function f in problem 1, explain why the conclusion of the intermediate value theorem is false if y = 5? f(1) = 2 and f(5) = 8 y = 5 is between f(1) and f(5). But the conclusion that there exists at least one number c such that f(c) = 5 is false. Looking at the graph there are no values of x where y = 5. Chunking: Problems 1-3 The purpose of these three problems is to show that the Intermediate Value Theorem conclusion might be true but the conclusion is only guaranteed if the function is continuous. In problems 1-3, the function f is not continuous and neither y= 5 not y = 3 are guaranteed but there is a number c where f(c) = 3.

For function f in problem 1, is there a value of x for which y = 3
For function f in problem 1, is there a value of x for which y = 3? Does this fact contradict the intermediate value theorem? Yes there is a value, x = 2, where f(2) = 3. This does not contradict the intermediate value theorem because the theorem only guarantees a value of c when the function is continuous, it does not rule out there might be a value of c such that f(c) = 3. Chunking: Problems 1-3 The purpose of these three problems is to show that the Intermediate Value Theorem conclusion might be true but the conclusion is only guaranteed if the function is continuous. In problems 1-3, the function f is not continuous and neither y= 5 not y = 3 are guaranteed but there is a number c where f(c) = 3.

The figure shows a function g defined on the closed interval [0, 5]
The figure shows a function g defined on the closed interval [0, 5]. Does the function g meet the hypotheses of the intermediate value theorem on that interval? Why or why not? Is there a value of x = c in [0, 5] for every value of k between g(0) and g(5) for which g(c) = k? Does this fact contradict the intermediate value theorem? How does the function g in problem 4 show the intermediate value theorem is not an “if and only if” theorem? Chunking: Problems 4-6 This is a great example where none of the values of g are skipped on the interval [1, 5]. Another of example of what the Intermediate Value Theorem does not say.

The figure shows a function g defined on the closed interval [0, 5]
The figure shows a function g defined on the closed interval [0, 5]. Does the function g meet the hypotheses of the intermediate value theorem on that interval? Why or why not? The function g does not meet the hypotheses of the intermediate value theorem on the interval [0. 5] because g is not continuous at x = 3. Chunking: Problems 4-6 This is a great example where none of the values of g are skipped on the interval [1, 5]. Another of example of what the Intermediate Value Theorem does not say.

Is there a value of x = c in [0, 5] for every value of k between g(0) and g(5) for which g(c) = k? Does this fact contradict the intermediate value theorem? Yes there is a value of x = c in [0, 5] for every value of k between g(0) and g(5) for which g(c) = k. This does not contradict the theorem because the theorem does not rule out the possibility that the conclusion could be true. Chunking: Problems 4-6 This is a great example where none of the values of g are skipped on the interval [1, 5]. Another of example of what the Intermediate Value Theorem does not say.

“if and only if” means the converse of the
How does the function g in problem show the intermediate value theorem is not an “if and only if” theorem? “if and only if” means the converse of the intermediate value theorem is true. But this example shows that even though there is a value of x = c in [0, 5] for every value of k between g(0) and g(5) for which g(c) = k, the function g is not continuous. Chunking: Problems 4-6 This is a great example where none of the values of g are skipped on the interval [1, 5]. Another of example of what the Intermediate Value Theorem does not say.

Intermediate Value Theorem

The Intermediate Value Theorem (IVT) says, roughly speaking, that a continuous function cannot skip values. Consider a plane that takes off and climbs from 0 to 10,000 meters in 20 minutes. The plan must reach every altitude between 0 and 10,000 meters during this 20-minute interval. Thus, at some moment, the plane’s altitude must have been exactly meters. Of course, this assumes that the plane’s motion is continuous, so its altitude cannot jump abruptly from, say, 8000 to 9000 meters.

To state this conclusion formally, let A(t) be the plane’s altitude at time t.
The IVT asserts that for every altitude M between 0 and 10,000, there is a time t0 between 0 and 20 such that A(t0) = M. In other words, the graph of A(t) must intersect the horizontal line y = M. By contrast, a discontinuous function can skip values.

Intermediate Value Theorem If f(x) is continuous on a closed interval [a, b] and f(a) ≠ f(b), then for every value M between f(a) and f(b), there exists at least one value 𝑐∈ 𝑎,𝑏 such that f(c) = M.

Example Prove that the equation sin x = 0.3 has at least one solution in the interval 0, 𝜋 2 .

The IVT can be used to show the existence of zeros of functions
The IVT can be used to show the existence of zeros of functions. If f(x) is continuous and takes on both positive and negative values – say f(a) < 0 and f(b) > 0 – then the IVT guarantees that f(c) = 0 for some c between a and b. Corollary to the IVT – If f(x) is continuous on [a, b] and if f(a) and f(b) are nonzero and have opposite signs, then f(x) has a zero in (a, b). We can locate zeros of functions to arbitrary accuracy using the Bisection Method.

Example Show that 𝑓 𝑥 = cos 2 𝑥 −2 sin 𝑥 4 has a zero in (0, 2). Then locate the zero more accurately using the Bisection Method.

Exploration: Mean Value Theorem
This exploration takes a graphical approach to the Mean Value Theorem (formally stated later). Later when differentiation techniques are learned this theorem will be revisited and algebraic solution can be found. This activity is also connected as a follow up of the Average Value Exploration day 15. All answers are approximate.

Draw the secant line and the tangent line.
For the function f graphed below, there is a value of x = c between 3 and 7 at which the tangent line to the graph of f is parallel to the secant line through (3, f(3)) and (7, f(7)). Draw the secant line and the tangent line. From the graph, c ≈ __________ Is f differentiable on (3, 7)? __________ Is f continuous on [3, 7]? __________ Chunking: Problem 1

Draw the secant line and the tangent line. From the graph, c ≈ 5.5
For the function f graphed below, there is a value of x = c between 3 and 7 at which the tangent line to the graph of f is parallel to the secant line through (3, f(3)) and (7, f(7)). Draw the secant line and the tangent line. From the graph, c ≈ 5.5 Is f differentiable on (3, 7)? Yes Is f continuous on [3, 7]? Yes Chunking: Problem 1

Draw the secant line and the tangent lines.
The function f in problem 1 has two values of x = c between x = 1 and x = 7 at which 𝑓 ′ 𝑐 equals the slope of the secant line through (1, f(1)) and (7, f(7)). That is, the tangent line is parallel to the secant line. Draw the secant line and the tangent lines. From the graph, c ≈ __________ and c ≈ __________ Is f differentiable on (1, 7)? Is f continuous on [1, 7]? Chunking: Problem 2 Key difference in problem 2: There is more than one point. So in the discussion you can bring out that there is at least one point. Also the notation for the derivative was used for the slope of the tangent line that is parallel to the secant line.

Draw the secant line and the tangent lines.
The function f in problem 1 has two values of x = c between x = 1 and x = 7 at which 𝑓 ′ 𝑐 equals the slope of the secant line through (1, f(1)) and (7, f(7)). That is, the tangent line is parallel to the secant line. Draw the secant line and the tangent lines. From the graph, c ≈ 5.6 and c ≈ 2.3 Is f differentiable on (1, 7)? Yes Is f continuous on [1, 7]? Yes Chunking: Problem 2

For the function g graphed below:
Draw a secant line through (1, g(1)) and (5, g(5)). Is g differentiable on (1, 5)? Is g continuous on [1, 5]? Tell why there is no value of x = c between x = 1 and x = 5 at which 𝑔 ′ 𝑐 equals the slope of the secant line draw in part (a). Chunking: Problems 3-4 Using the Geogebra file Graphs for Mean Value in the Addendum select function g(x), segment h and line i then drag point K to show there is no tangent parallel to the secant in the interval.

For the function g graphed below:
Draw a secant line through (1, g(1)) and (5, g(5)). Is g differentiable on (1, 5)? No Is g continuous on [1, 5]? Yes Tell why there is no value of x = c between x = 1 and x = 5 at which 𝑔 ′ 𝑐 equals the slope of the secant line draw in part (a). The slope of the secant line is about ½ and along the left branch at x = 1, 𝑔 ′ 𝑥 ≈ and 𝑔 ′ 𝑥 is increasing so there is no point along the left branch where 𝑔 ′ 𝑐 ≈ On the right branch the rate of change is negative. Chunking: Problems 3-4 Ask questions to lead students to the real issue here that g is not differentiable on (1, 5). Ask the question: So what changed from problems 1 and 2 to problem 3? What caused there to be no value of x = c between x = 1 and x = 5 at which g’(c) equals the slope of the secant line?

The function g from problem 3 does have a value x = c in (1, 4) for which 𝑔 ′ 𝑐 equals the slope of the secant line through (1, g(1)) and (4, g(4)). Draw the secant line and the tangent line below. From the graph, c ≈ __________. Is g differentiable on (1, 4)? Is g continuous on [1, 4]?

Draw the secant line and the tangent line below.
The function g from problem 3 does have a value x = c in (1, 4) for which 𝑔 ′ 𝑐 equals the slope of the secant line through (1, g(1)) and (4, g(4)). Draw the secant line and the tangent line below. From the graph, c ≈ 3. Is g differentiable on (1, 4)? Yes Is g continuous on [1, 4]? Yes Emphasize here that there is continuity and differentiability.

For the function h graphed below:
Draw a secant line through (5, h(5)) and (7, h(7)). Is h differentiable on (5, 7)? Is h continuous on [5, 7]? Tell why there is no value of x = c in (5, 7) at which ℎ ′ 𝑐 equals the slope of the secant line drawn in part (a). Chunking: Problems 5-7

For the function h graphed below:
Draw a secant line through (5, h(5)) and (7, h(7)). Is h differentiable on (5, 7)? Yes Is h continuous on [5, 7]? No Tell why there is no value of x = c in (5, 7) at which ℎ ′ 𝑐 equals the slope of the secant line drawn in part (a). The slope of the secant line is about -1. The values of ℎ ′ 𝑐 are all positive on (5, 7). Chunking: Problems 5-7

The graph below is the function h from problem 5.
Draw a secant line through (5, h(5)) and (9, h(9)). Is h differentiable on (5, 9)? Is h continuous on [5, 9]? There is a value of x = c in (5, 9) where ℎ ′ 𝑐 equals the slope of the secant line drawn in part (a). Draw the tangent line. Estimate the value of c. c ≈ __________

Draw a secant line through (5, h(5)) and (9, h(9)). Is h differentiable on (5, 9)? Yes Is h continuous on [5, 9]? No There is a value of x = c in (5, 9) where ℎ ′ 𝑐 equals the slope of the secant line drawn in part (a). Draw the tangent line. Estimate the value of c. c ≈ 6 Key issue here is the fact that the function h is not continuous but h is differentiable.

Draw a secant line through (0, h(0)) and (9, h(9)). Show that there are two points x = c in (0, 9) where ℎ ′ 𝑐 equals the slope of the secant line drawn in part (a). Draw the tangent lines and estimate the values of c. c ≈ __________ and c ≈ __________ Is h differentiable on (0, 9)? Is h continuous on [0, 9]?

Draw a secant line through (0, h(0)) and (9, h(9)). Show that there are two points x = c in (0, 9) where ℎ ′ 𝑐 equals the slope of the secant line drawn in part (a). Draw the tangent lines and estimate the values of c. c ≈ 3 and c ≈ 5.5 Is h differentiable on (0, 9)? No Is h continuous on [0, 9]? No

For which problem(s) are: the hypotheses and conclusion true?
The Mean Value Theorem states: If f is differentiable on (a, b) and f is continuous on [a, b], then there is a number c (the mean value) in (a, b) such that 𝑓 ′ 𝑐 = 𝑓 𝑏 −𝑓 𝑎 𝑏−𝑎 . The slope of the tangent line equals the slope of the secant line. For which problem(s) are: the hypotheses and conclusion true? the hypotheses and conclusion not true? the conclusion is true but not the hypotheses? Chunking: Problem 8

For which problem(s) are:
The Mean Value Theorem states: If f is differentiable on (a, b) and f is continuous on [a, b], then there is a number c (the mean value) in (a, b) such that 𝑓 ′ 𝑐 = 𝑓 𝑏 −𝑓 𝑎 𝑏−𝑎 . The slope of the tangent line equals the slope of the secant line. For which problem(s) are: the hypotheses and conclusion true? 1, 2, and 4 the hypotheses and conclusion not true? 3 and 5 the conclusion is true but not the hypotheses? 6 and 7 Chunking: Problem 8 Make sure the students understand the formal statement of the Mean Value Theorem and its relationship to the previous problems.

Summarize what you learned about the mean value theorem.

Summarize what you learned about the mean value theorem.
One of the big ideas is that the hypotheses guarantee the conclusion that instantaneous rate of change must equal the average rate of change, but the conclusion can be true even when the hypotheses is not true.

IVT and MVT Review

The IVT can be used to show the existence of zeros of functions
The IVT can be used to show the existence of zeros of functions. If f(x) is continuous and takes on both positive and negative values – say f(a) < 0 and f(b) > 0 – then the IVT guarantees that f(c) = 0 for some c between a and b. Corollary to the IVT – If f(x) is continuous on [a, b] and if f(a) and f(b) are nonzero and have opposite signs, then f(x) has a zero in (a, b). We can locate zeros of functions to arbitrary accuracy using the Bisection Method.

For which problem(s) are: the hypotheses and conclusion true?
The Mean Value Theorem states: If f is differentiable on (a, b) and f is continuous on [a, b], then there is a number c (the mean value) in (a, b) such that 𝑓 ′ 𝑐 = 𝑓 𝑏 −𝑓 𝑎 𝑏−𝑎 . The slope of the tangent line equals the slope of the secant line. For which problem(s) are: the hypotheses and conclusion true? the hypotheses and conclusion not true? the conclusion is true but not the hypotheses? Chunking: Problem 8

Use the IVT to show that 𝑓 𝑥 = 𝑥 3 +𝑥 takes on the value 9 for some x in [1,2 ].
Show that 𝑓 𝑥 =𝑥− cos 𝑥 has a zero in [0, 1]. Apply the Bisection Method twice to narrow the window. Find a point c satisfying the conclusion of the MVT for 𝑦= 𝑥 3 on the interval [-4, 5].

IVT and MVT Quiz

Fill in the blank. IVT: If f is ____________________ on a closed interval [a, b] and k is a number between _____ and _____ then there is at least one number c in [a, b] such that f(c) = k. Corollary to the IVT: If f(x) is ____________________ on [a, b] and if f(a) and f(b) have ________________ signs, then f(x) has a __________ in (a, b). MVT: If f is ____________________ on (a, b) and f is ____________________ on [a, b], then there is a number c (the mean value) in (a, b) such that 𝑓 ′ 𝑐 =______________.

Use the IVT to show that 𝑓 𝑥 = 𝑥 2 −𝑥 takes on the value 6 for some x in [2, 5].
Show that 𝑓 𝑥 = 𝑥 7 +3𝑥−10 has a zero in [1, 2]. Apply the Bisection Method twice to narrow the window. Find a point c satisfying the conclusion of the MVT for 𝑦= 𝑥 on the interval [9, 25].

Limits at Infinity

We have considered limits as x approaches a number c.
It is also important to consider limits where x approaches ∞ or -∞, which we refer to as limits at infinity. In applications, limits at infinity arise naturally when we describe the “long-term” behavior of a system as shown in the graph.

In either case, the line y = L is called a horizontal asymptote.
The notation x ⟶ ∞ indicates that x increases without bound, and x ⟶ -∞ indicates that x decreases (through negative values) without bound. lim 𝑥→∞ 𝑓 𝑥 =𝐿 if f(x) gets closer and closer to L as x ⟶ ∞ lim 𝑥→−∞ 𝑓 𝑥 =𝐿 if f(x) gets closer and closer to L as x ⟶ -∞ In either case, the line y = L is called a horizontal asymptote. Infinite limits describe the asymptotic behavior of a function, which is the behavior of the graph as we move out to the right or left.

Example Discuss the asymptotic behavior in the figure below.

Example Discuss the asymptotic behavior in the figure below. There is a horizontal asymptote of y = 7 as we move to the right. lim 𝑥→∞ 𝑔 𝑥 =7 There is a horizontal asymptote of y = 3 as we move to the left. lim 𝑥→−∞ 𝑔 𝑥 =3

A function may approach an infinite limit as x ⟶ ±∞.
Similar notation is used if f(x) approaches -∞

The graph of 𝑦= 𝑒 𝑥 is shown. What are the following limits?

Limits at infinity do not always exist.
For example, 𝑓 𝑥 = sin 𝑥 oscillates indefinitely so lim 𝑥→±∞ 𝑓 𝑥 do not exist.

The limits at infinity of the power functions 𝑓 𝑥 = 𝑥 𝑛 are easily determined

Theorem For all n > 0, lim 𝑥→∞ 𝑥 𝑛 =∞ and lim 𝑥→∞ 𝑥 −𝑛 = lim 𝑥→∞ 1 𝑥 𝑛 =0 If n is a whole number, lim 𝑥→−∞ 𝑥 𝑛 = ∞ if 𝑛 is even −∞ if 𝑛 is odd lim 𝑥→−∞ 𝑥 −𝑛 = lim 𝑥→−∞ 1 𝑥 𝑛 =0

Theorem The asymptotic behavior of a rational function depends only on the leading terms of its numerator and denominator. If 𝑎 𝑛 , 𝑏 𝑛 ≠0, then lim 𝑥→±∞ 𝑎 𝑛 𝑥 𝑛 + 𝑎 𝑛−1 𝑥 𝑛−1 +…+ 𝑎 0 𝑏 𝑚 𝑥 𝑚 + 𝑏 𝑚−1 𝑥 𝑚−1 +…+ 𝑏 0 = 𝑎 𝑛 𝑏 𝑚 lim 𝑥→±∞ 𝑥 𝑛−𝑚 Proof of this theorem is in the textbook.

Use the theorem to evaluate the following limits.

Given the graph of 𝑓 𝑥 = 𝑎 𝑛 𝑥 𝑛 + 𝑎 𝑛−1 𝑥 𝑛−1 +…+ 𝑎 0 𝑏 𝑚 𝑥 𝑚 + 𝑏 𝑚−1 𝑥 𝑚−1 +…+ 𝑏 0 behaves as the graph of 𝑔 𝑥 = 𝑎 𝑥 𝑛 𝑏 𝑥 𝑚 = 𝑎 𝑏 𝑥 𝑛−𝑚 : If n = m, lim 𝑥→±∞ 𝑓 𝑥 = 𝑎 𝑏 If n < m, lim 𝑥→±∞ 𝑓 𝑥 =0 If n > m, lim 𝑥→±∞ 𝑓 𝑥 =±∞ Think about the end behaviors of the graph of g(x) to determine if it is +∞ or −∞

Limits and Continuity AP Calculus AB.

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Limits and Continuity AP Calculus AB.

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