3B MAS 4. Functions

Limit of a Function Graphically the limiting value of a function f(x) as x gets closer and closer to a certain value (say 'a') is obtained by moving along the curve from both sides of 'a' as x moves toward 'a'. The limiting value of f(x) as x gets closer and closer to 'a' is denoted by

Right/Left Hand Limits As x moves towards 'a' from right (left) hand side, the limiting value of f(x) is denoted by

Limiting and Functional Value If both sides limits are equal, Otherwise, does not exist. Note that may not equal to f(a)

a f(x) x y f(a) Limiting Value

Example 1 Find the limit of the f(x) as x approaches a for the following functions. (a) a

Example 1 Find the limit of the f(x) as x approaches a for the following functions. (a) a The limit does not exist as the function is not defined 'near' a.

Example 1 (cont'd) (b) a

Example 1 (cont'd) (b) a The limit does not exist as the left side limit is not the same as right side limit.

Example 1 (cont'd) (c) f(a) a

Example 1 (cont'd) (c) f(a) a The limit exists but it does not equal to f(a).

Example 1 (cont'd) (d) a f(a)

Example 1 (cont'd) (d) The limit exists and it equals to f(a). a f(a)

Evaluating Limits If f(x) is not broken at 'a', use direct substitution to evaluate its limit as x approaches 'a' Otherwise, find the left side and right side limits and check if they are equal.

Example 2 Evaluate the following limits if they exist. (a) f(x) = 2x – 5 as x  1 f(x) is not broken at x = 1, so use direct sub.

Example 2 (cont'd) (b) f(x) = ln x as x  0 f(0) is not defined. So consider limit from both sides. But f(x) is not defined for x < 0. So the limit does not exist.

Example 2 (c) f(x) = 1/(x – 2) as x  2 f(2) is not defined. So consider limit from both sides. Since the left side limit does not equal to the right side limit, the limit of the f(x) as x approaches 2 does not exist.

Example 2 (cont'd) (d) f(x) = (x – 1)/(x 2 – 1) as x  1 f(x) = (x – 1)/(x + 1)(x – 1) = 1/(x + 1) 1/(x + 1) is not broken at x = 1, so use direct sub.

Example 2 (cont'd)

Limits to Infinity If f(x) = x + c, f(x)   as x   (note that x is the dominant term) If f(x) = 1/x, f(x)  0 as x   If f(x) = ax 2 + bx + c, ax 2 is the dominant term as x  

Example 3 Find the limit of f(x) as x   (if they exist) for:

Example 3 (cont'd)

Trigonometric Limits

Example 4 Find the following limits.

x C B AO r Consider the relationship between the areas  OAC, sector OAC, and  OAB

x C B AO r Area of  OAC = r 2 sin x / 2 Area of sector OAC = r 2 x / 2 Area of  OAB = r 2 tan x / 2 Example 4 (cont'd)

x C B AO r Area of  OAC = r 2 sin x / 2 Area of sector OAC = r 2 x / 2 Area of  OAB = r 2 tan x / 2 So (size of areas) r 2 sin x / 2 sin x / x > cos x Take limit as x  0 to get That means sin x  x as x  0

Example 4 (cont'd)

Continuity Graphically a graph is continuous at x = a if it is not broken (disconnected) at that point. Algebraically the limit of the function from both sides of 'a' must equal to f(a).

Example 5 The following functions are not continuous at x = a. a a f(a) a Why?

Example 6 The following functions are continuous at x = a. a f(a) a

Example 7 Determine if the given function is continuous at the given point. (a) f(x) = | x – 2 | at x = 2 (b) f(x) =  xat x = 0 (c) f(x) = 1 / (x + 3)at x = -3

Example 7 (cont'd)

Example 8 Given that f(x) is continuous over the set of all real numbers, find the values of a and b.

Only need to consider the junctions (x = -1 and x = 2) Example 8 (cont'd)

Differentiability Graphical approach: A function f(x) is said to be differentiable at x = a if there is no 'corner' or 'vertical tangency' at that point. A function must be continuous (but not sufficient) in order that it may be differentiable at that point.

Example 9 The following functions are not differentiable at x = a. (a) a f(a) Corner at x = a

Example 9 (cont'd) (b) Vertical tangency at x = 1

Example 9 (cont'd) (c) Not continuous (not even defined) at x = -2

Example 10 The following functions are differentiable everywhere. (a)

Example 10 (cont'd) (b)

Example 10 (cont'd) (c)

Derivative of a Function A function is differentiable at a point if it is continuous (not broken), smooth (no corner) and not vertical (no vertical tangency) at that point. Its derivative is given by (First Principle)

f(x) f(x+h) x x+h P Q Differentiability (cont'd)

The gradient of PQ is given by As Q moves closer and closer to P (i.e. as h tends to 0), the limiting value of the gradient of PQ (i.e. the derivative of f(x) at x) becomes the tangent at P.

Differentiability (cont'd) The derivative of a function y = f(x) is denoted by It also represents the rate of change of y with respect to x.

Example 11 (a)Find the gradient function of y = 2x 2 using first principle. Find also the gradient at the point (3, 18). (b)Use the definition (first principle) to find the derivative of ln x and hence find the derivative of e x.

Example 11 (cont'd)

Example 12 Find the derivative of the following functions from first principles. (a) f(x) = 1/x (b) f(x) =  x (c) f(x) = x n

Example 12 (cont'd)

Concavity If f(x) opens downward, it is said to be concave down If f(x) opens upwards, it is concave up concave down concave up f '(x):

Point of Inflection Points of inflection: points where the curve changes from concave up to concave down or concave down to concave up point of inflection f '(x): maximum

Horizontal Inflection Horizontal inflection: a point of inflection where the graph is momentarily horizontal, dy/dx = 0 horizontal inflection -ve +ve

Stationary Points Turning points: max and min Stationary points: max, min and horizontal inflection dy/dx = 0

Example 13 Given the f(x) graph below draw f '(x).

f '(x) f (x) a b c f '(x) changes as below: x < a: +ve, but  x = a: f '(a) = 0 local max a < x < b: -ve, , then  (less –ve) point of inflection Example 13 (cont'd)

f '(x) f (x) a b c x = b: f '(b) = 0 local min b < x < c: +ve, , then  point of inflection x = c: f '(c) = 0 global max x > c: -ve,  Example 13 (cont'd)

Noteworthy Features Min TP: dy/dx = 0, sign change –ve, 0, +ve Max TP: dy/dx = 0, sign change +ve, 0, –ve Horizontal inflection: dy/dx = 0, +ve, 0, +ve or –ve, 0, –ve, (i.e. no sign change) Point of inflection: d 2 y/dx 2 = 0, (dy/dx is a max/min),

Example 14 Graph the following function and its derivative. Use your graphs to locate the stationary points and points of inflection on y = x 4 /4 – 4x 3 /3 – 7x 2 /2 + 10x + 5 and determine the nature of each.

y = x 4 /4 – 4x 3 /3 – 7x 2 /2 + 10x + 5 dy/dx = x 3 – 4x 2 – 7x + 10 TP  dy/dx = 0 So x 3 – 4x 2 – 7x + 10 = 0 i.e. (x + 2)(x – 1)(x – 5) = 0  x = -2, 1 or 5 When x = -2, y = -43/3 When x = 1, y = 125/12 When x = 5, y = -515/12 Example 14 (cont'd)

f(x) f '(x) min max minPoI

Piecewise Defined Functions A piecewise defined function has different formulas for different parts of its domain. At junction a filled circle indicates that a point actually exists there, whereas an empty circle shows a discontinuous point.

Example 15 Given the function below (a)Find f(-2), f(1) and f(2) (b)Graph f and determine whether f is continuous at x = 0 and x = 2.

(a)f(-2) = 1/(-2) – 1 = -3/2 f(1) = 1 2 = 1 f(2) = = 3 Example 15 (cont'd)

Example 16 Graph y = | x 2 – 6x + 8 | and determine whether the function is continuous at x = 2 and x = 4.

Example 16 (cont'd) From graph, the function is continuous at x = 2 and x = 4.

The Sign Function It can be considered as a logical function (especially in computer science) It extracts the sign of the function It returns 1 if f(x) is positive, 0 if f(x) equals to 0 and –1 if f(x) is negative.

Sign Function (cont'd)

Example 17 Graph (a)y = sgn (x/|x|) (b)y = sgn (x 2 – 1)

(a) y = sgn (x/|x|) Example 17 (cont'd) Not continuous at x = 0

(b) y = sgn (x 2 – 1) Example 17 (cont'd) Not continuous at x = -1 and x = 1

Greatest Integer Function Also known as floor function Defined as the greatest integer less than or equal to the number That is, it rounds any number down to the nearest integer Symbol: int [x] or int [4.2] = = 4 int [-2.1] = = -3

Greatest Integer Function (cont'd) Not continuous at all integers.

Example 18 Graph the following functions: (a)int [2x – 1] (b)int [x 2 ]

(a) int [2x –1] Consider 2x – 1 = n where n is an integer x = (n + 1) / 2 So the 'breaking points' are steps of half of an integer Example 18 (cont'd)

(b) int [x 2 ] Consider x 2 = n where n is a positive integer x =  n So the breaking points are  square root of +ve integers Example 18 (cont'd)

Rules of Differentiation (Review)

Example 19 Find the derivative of the following functions:

Example 19 (cont'd)

Differentiating e f(x) and ln f(x)

Example 20 Differentiate the following with respect to x:

Example 20 (cont'd)

Example 21 Differentiate

Example 21 (cont'd)

Differentiability (Revisit) Graphical approach: continuous, no corner, no vertical tangency Algebraical approach:

Example 22 Determine if the following functions are differentiable at the indicated points. (a) y = 1 / (x + 1)at x = -1 (b) y = | x + 1 |at x = -1 (c) f(x) = -6x + 5for x < 3 = -x 2 – 4 for x  3at x = 3

(a) Let f(x) = 1 / (x + 1) f(-1) is not defined f(x) is not continuous at x = -1 f(x) is not differentiable at x = -1 Example 22 (cont'd)

Example 23 Find the value of a and b so that f(x) = 3x + 1for x < 1 = x 2 + ax + bfor x  1 is continuous and differentiable everywhere.

Possible discontinuity and non-differentiability at x = 1 continuous at x = 1 if4 = 1 + a + b i.e.a + b = 3 f ’(x) = 3for x 1 differentiable at x = 1 if3 = 2 + a Soa = 1and b = 2 Example 23 (cont'd)

Riemann Sums To find an approximate area under a curve between two x values [a, b] The area is divided into n rectangles of equal width So the width  x = (b – a)/n There are many ways to find the height h of each rectangle (see later) Then the required area A =  h  x over the interval [a, b]

Example 24 The shaded area below shows the exact area under the curve f(x) = x 3 – 3x in the interval [0, 3] y = x 3 – 3x Actual area = 17.25

Example 25 (Example 24) Consider n = 5,  x = (3 – 0)/5 = 0.6 and h = left endpoint  x = (3 – 0)/5 = 0.6 Left endpoint for the 3 rd rectangle

Example 25 (cont'd) f(x 1 )= (0) 3 – 3(0) = 8 f(x 2 )= (0.6) 3 – 3(0.6) = f(x 3 )= (1.2) 3 – 3(1.2) = f(x 4 )= (1.8) 3 – 3(1.8) = f(x 5 )= (2.4) 3 – 3(2.4) = 4.544

Example 26 (Example 24) Consider n = 5,  x = (3 – 0)/5 = 0.6 and h = right endpoint Right endpoint for the 3 rd rectangle

Example 26 (cont'd) f(x 1 )= (0.6) 3 – 3(0.6) = f(x 2 )= (1.2) 3 – 3(1.2) = f(x 3 )= (1.8) 3 – 3(1.8) = f(x 4 )= (2.4) 3 – 3(2.4) = f(x 5 )= (3.0) 3 – 3(3.0) = 8

Example 27 (Example 24) Consider n = 5,  x = (3 – 0)/5 = 0.6 and h = minimum point Minimum point for the 4 th rectangle

Example 27 (cont'd) f(x 1 )= (0.6) 3 – 3(0.6) = f(x 2 )= (1.2) 3 – 3(1.2) = f(x 3 )= (1.8) 3 – 3(1.8) = f(x 4 )= (2.0) 3 – 3(2.0) = 4 f(x 5 )= (2.4) 3 – 3(2.4) = 4.544

Example 28 (Example 24) Consider n = 5,  x = (3 – 0)/5 = 0.6 and h = maximum point Maximum point for the 4 th rectangle

Example 28 (cont'd) f(x 1 )= (0) 3 – 3(0) = 8 f(x 2 )= (0.6) 3 – 3(0.6) = f(x 3 )= (1.2) 3 – 3(1.2) = f(x 4 )= (2.4) 3 – 3(2.4) = f(x 5 )= (3.0) 3 – 3(3.0) = 8

Example 29 (Example 24) Consider n = 5,  x = (3 – 0)/5 = 0.6 and h = midpoint Midpoint for the 3 rd rectangle

Example 29 (cont'd) f(x 1 )= (0.3) 3 – 3(0.3) = f(x 2 )= (0.9) 3 – 3(0.9) = f(x 3 )= (1.5) 3 – 3(1.5) = f(x 4 )= (2.1) 3 – 3(2.1) = f(x 5 )= (2.7) 3 – 3(2.7) = 5.813

A Better Approximation Due to the use of h (left, right, mid, min and max), the rectangles do not truly represent the area under the curve for each strip If n (number of rectangles) increases, the error decreases

Example 30 n = 20,  x = (3 – 0)/20 = 0.15, h = midpoint Actual area = 17.25

Example 31 n = 100,  x = (3 – 0)/100 = 0.03, h = midpoint Actual area = 17.25

Limit of a Sum The more rectangles, the greater accuracy So the actual area A is given by This is written as

Integration  f(x)dx represents the area under the curve f(x)

Example 32 b34567x Area Evaluate the area under the curve y = 3 from x = 2 to x = b by completing the following table. Hence give an answer for where k is a constant

Example 32 (cont'd) b34567b Area (b – 2)

Evaluate the area under the curve y = 2x from x = 0 to x = b by completing the following table. Hence give an answer for where k is a constant Example 33 b01234b Area

Example 33 (cont'd) b01234x Area b  2b/2 = b 2

Given the areas under the curve y = x 2 from x = 0 to x = 4 in the following table, find the area when x = b Hence give an answer for Example 34 b01234b Area01/38/3964/3

Example 34 (cont'd) b01234b Area01/38/3964/3b 3 /3