Extreme Values of Functions

What you’ll learn… . Absolute (global) extreme values Local (relative) extreme values Finding extreme values Inflection point and concavity Graph a function Why? Finding maximum and minimum values of functions, called optimization, is an important issue in real-world problems .

The Extreme Value Theorem If f is continuous on a closed interval [a,b], then f has both a maximum value and a minimum value on the interval. 2 conditions for f: continuous & closed interval If either condition does not exist the E.V.T. does not apply.

First Derivative Test for Local Extrema

First Derivative Test for Local Extrema The following test applies to a continuous function f(x). At a critical point c: If f ‘ changes signs from positive to negative at c, then f has a local maximum value at c. If f ‘ changes signs from negative to positive at c, then f has a local minimum value at c. If f ‘ does not change signs at c, then f has no local extreme value at c.

First Derivative Test for Local Extrema The following test applies to a continuous function f(x). At a left endpoint a: If f ‘ < 0 for x > a, then f has a local maximum value at a. If f ‘ > 0 for x > a, then f has a local minimum value at a. At a right endpoint b: If f ‘ < 0 for x < b, then f has a local minimum value at b. If f ‘ > 0 for x < b, then f has a local maximum value at b.

Example Using the First Derivative Test 1. Since f is differentiable for all real numbers, the only critical points are the zeros of f ‘. l l inc. dec. inc. Local Maximum at x = – 3 Local Minimum at x = 3 The range of f (x) is No absolute extrema.

Concavity : Compare f(x) and g(x) Both are increasing functions but they don’t look quite the same.

Concavity Concavity Test

Example Determining Concavity 2.

Inflection Points An inflection point is a point on the graph where the concavity changes from upward to downward or downward to upward. This means that if f ’’(x) exists in a neighborhood of an inflection point, then it must change sign at that point. Theorem 1. If y = f (x) is continuous on (a,b) and has an inflection point at x = c, then either f ’’(c) = 0 or f ’’(c) does not exist.

Point of Inflection . Points of Inflection:

Example 2 Find the inflection point(s) of f(x) = x3 – 9x2 +24x -10 F’(x) = 3x2 – 18x + 24 F’’(x) = 6x – 18 = 0 6(x-3) = 0 x = 3 Note: It’s important to do this test because the second derivative must change sign in order for the graph to have an inflection point. x 2 3 4 F’’ -- + Concave down Concave up Therefore 3 is the infection point of f(x)

Example 3: A special case Find the inflection point(s) of f(x) = x4 F’(x) = 4x3 F’’(x) = 12x2 = 0 x = 0 x -1 1 F’’ + + Concave up Concave up Therefore 0 is not the inflection point of f(x) There is no inflection point for this graph

Example 4 Find the inflection point(s) of f(x) = ln(x2 - 2x + 5) x = -1 and x =3 x -2 -1 3 4 F’’ - + - Concave up Concave down Concave down Therefore there are two inflection points at x= -1 and x=3

Second Derivative Test for Local Extrema Example Using the Second Derivative Test 4. Local Maximum at: Local Minimum at:

Learning about Functions from Derivatives

Use the following function to find each of the following. Local min. at x = 3 Identify where the extrema of f occur. Find the intervals on which f is increasing and decreasing. Find where the graph of f is concave up or down. Sketch a possible graph for f. l l Dec Dec Inc l l Up Down Up

With today technology, graphing calculator and computer can produce graphs. However, important points on a plot may be difficult to identify. Therefore, it’s useful to learn how to sketch a graph by hand.

Curve Stretching Analyze f(x). Find the domain and intercepts. (Set x=0, solve for f(x); set f(x)=0, solve for x). Analyze f’(x): Find critical values. Determine increasing and decreasing intervals as well as local maximum and/or minimum. (set f’(x)=0). Analyze f’’(x): Find inflection point. Determine the intervals on which the graph is concave upward and concave downward. (set f’’(x)=0). Plot additional points as needed and sketch the graph.

Example Sketch f(x) = x4 + 4x3 by hand Step 1: Domain: (-∞,∞) X: intercept: x4 + 4 x3 = 0 x3 (x+4) = 0, so x=0 or x = -4 Y: intercept: f(0) = 0 Step 2: f’(x) = 4x3 + 12x2 = 0 4x2 (x+3) = 0 so x= 0 or x=-3 both critical v. Test numbers on the left and on the right of 0 and -3, we see that -3 is a local minimum. Also, f(x) is decreasing on (- ∞, -3) and increasing on (-3, ∞). Step 3: f’’(x) = 12x2 + 24x = 0 12x(x+2) = 0 so x = 0 or x = -2 Test numbers on the left and on the right of -2 and 0, we see that both of them are inflection points. Also, the graph is concave upward on (- ∞, -2), concave downward on (-2,0), and concave upward on (0, ∞)

Continue: Sketch f(x) = x4 + 4x3 Note -4 x-int -3 -27 min -2 -16 Inflection point x-int, y-int (- ∞, -3) decreasing (-3, ∞) increasing (- ∞, -2) concave up (-2,0) concave down (0, ∞)

Example Sketch f(x) = 3x2/3 - x by hand Step 1: Domain: (-∞,∞) X: intercept: 3x2/3 - x = 0 x (3x-1/3 - 1) = 0, so x=0 or 3x-1/3 – 1 = 0 x-1/3 = 1/3, (x-1/3)-3 = (1/3)-3 , so x = 27 Y: intercept: f(0) = 0 Step 2: f’(x) = 2x-1/3 -1 = 0 x-1/3 = 1/2, (x-1/3)-3 = (1/2)-3 , so x= 8 Also, f’(x) is discontinuous at 0. Test numbers on the left and on the right of 0 and 8, we see that 0 is a min and 8 is a max. Also, f(x) is decreasing on (- ∞, 0) and (8, ∞) and increasing on (0,8) Step 3: f’’(x) = (-2/3) x-4/3 = 0 x-4/3 = 0; so x = 0 F’’ is also discontinuous at 0. Test numbers on the left and on the right of 0, we see that there is no inflection point. The graph is concave downward on (- ∞, 0) and on (0, ∞)

Continue: Sketch f(x) = 3x2/3 - x Note X-int, y-int, min 8 4 max 27 X-int May need to add more points on the left of 0 to have a better graph (- ∞, 0) Decreasing Concave down (0,8) increasing (8, ∞) decreasing (0, ∞)

Quick Quiz

Quick Quiz