   Chapter 3.1, Problem 57E

Chapter
Section
Textbook Problem

# If a and b are positive numbers, find the maximum value of f ( x ) = x a ( 1 − x ) b ,    0 ≤ x ≤ 1.

To determine

To find:

The maximum value of f(x) on the given interval.

Explanation

1) Concept:

Use the closed interval method to find the maximum value of the given function.

The Closed Interval Method:

To find the absolute maximum and minimum values of a continuous function f on a closed interval a, b:

i. Find the values of f at the critical numbers of f in a, b.

ii. Find the values of f at the end points of the interval .

iii. The largest of the values from step (i) and (ii) is the absolute maximum value; the smallest of these values is the absolute minimum value.

2) Given:

fx=xa(1-x)b, 0 x 1 and a and b are positive numbers.

3) Calculation:

Since f(x) is continuous on 0, 1, use the Closed Interval Method to find the maximum value of the given function.

Differentiate f(x) with respect to x, and then find the values of x, where f'x=0 and f'x doesn’t exist.

By using the chain rule of derivative,

f'x=xa*b1-xb-1*(-1)+1-xb* axa-1

f'x=xa-11-xb-1a1-x-bx

f'x=xa-11-xb-1a-ax-bx

f'x=xa-11-xb-1a-x(a+b)

Now set f'x =0, and solve for x.

0=xa-11-xb-1a-x(a+b)

By using zero product property,

xa-1=0 or 1-xb-1 or a-x(a+b)=0

Solve for x.

x=0, 1 and a-x(a+b)=0

Hence,

x =0, 1, and aa+b

f'x exists for all x

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