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Chapter 1: Q. 52 (page 121)

Use the Extreme Value Theorem to show that each function f has both a maximum and a minimum value on [a, b]. Then use a graphing utility to approximate values M and m in [a, b] at which f has a maximum and a minimum, respectively. You may assume that these functions are continuous everywhere.

f(x)=3−2x2+x3,[a,b]=[−1,2]

Short Answer

Expert verified

The extreme value function guarantees that the function will not attain global maximum or minimum values on the given interval.

Step by step solution

01

Step 1. Given information.

We have been given a function and an interval as:

f(x)=3−2x2+x3,[a,b]=[−1,2]

We have to show that this function f has both a maximum and a minimum value on [a, b] using the Extreme Value Theorem.

Also, we have to find approximate values M and m in [a, b] at which f has a maximum and a minimum, respectively, using a graphing utility.

02

Step 2. Apply the Extreme Value Theorem 

limx→−1 f(x)=limx→−1 3−2x2+x3=3−2(−1)2+(−1)3=3−2−1=3−3=0limx→2 f(x)=limx→2 3−2x2+x3==3−222+23=3−2(4)+8=3−8+8=3

The extreme value function guarantees that the function will not attain global maximum or minimum values on the interval [-1,2].

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Most popular questions from this chapter

True/False: Determine whether each of the statements that follow is true or false. If a statement is true, explain why. If a statement is false, provide a counterexample.

(a) True or False: For limx→cf(x) to be defined, the function f must be defined at x = c.

(b) True or False: We can calculate a limit of the form limx→cf(x) simply by finding f(c).

(c) True or False: If limx→cf(x)=10, then f(c) = 10.

(d) True or False: If f(c) = 10, then limx→cf(x)=10.

(e) True or False: A function can approach more than one limit as x approaches c.

(f) True or False: If limx→4f(x)=10, then we can make f(x) as close to 4 as we like by choosing values of x sufficiently close to 10.

(g) True or False: If limx→6f(x)=∞, then we can make f(x) as large as we like by choosing values of x sufficiently close to 6.

(h) True or False: If limx→∞f(x)=100, then we can find values of f(x) between 99.9 and 100.1 by choosing values of x that are sufficiently large.

Sketch a labeled graph of a function that fails to satisfy the hypothesis of the Intermediate Value Theorem, and illustrate on your graph that the conclusion of the Intermediate Value Theorem does not necessarily hold.

Describe the punctured interval around x=2that has a radius of 3 and the punctured interval aroundx=4 that has a radius of 0.25.

Determine whether each of the statements that follow is true or false. If a statement is true, explain why. If a statement is false, provide a counterexample.

(a) A limit exists if there is some real number that it is equal to.

(b) The limit of fxas x→cis the value fc.

(c) The limit of fxas x→cmight exist even if the value of fcdoes not.

(d) The two-sided limit of fxas x→cexists if and only if the left and right limits of fxexists as x→c.

(e) If the graph of fhas a vertical asymptote at x=5, then limx→5fx=∞.

(f) If limx→5fx=∞, then the graph of fhas a vertical asymptote at x=5.

(g) If limx→2fx=∞, then the graph of fhas a horizontal asymptote at x=2.

(h) Iflimx→∞fx=2, then the graph offhas a horizontal asymptote aty=2.

Use the Extreme Value Theorem to show that each function f has both a maximum and a minimum value on [a, b]. Then use a graphing utility to approximate values M and m in [a, b] at which f has a maximum and a minimum, respectively. You may assume that these functions are continuous everywhere.

f(x)=x4−3x2−2,[a,b]=[−2,2]
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