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Chapter 5: Q 2TB. (page 489)

Bounding the second derivative: Consider the function f(x)=x2(x-3).

(a) To the nearest tenths place, find the smallest number Mso that f''(x)鈮�Mfor all x鈭�[0,2].

(b) To the nearest tenths place, find the smallest number Mso that f''(x)鈮�Mfor all x鈭�[0,4].

Short Answer

Expert verified

Part (a) The smallest value ofMis6for all values x鈭�[0,2].

Part (b) The smallest value ofMis6for all values x鈭�[0,4].

Step by step solution

01

Part (a) Step 1: Given Information

Consider the function.

f(x)=x2(x-3)

02

Part (a) Step 2: Simplify the given function and then determine the second derivative.

Simplify the given function.

f(x)=x2(x-3)=x3-3x2

The first derivative is as follows:

f'(x)=3x2-6x

The second derivative is as follows:

f''(x)=6x-6

03

Part (a) Step 3: Determine the value of M such that f''(x)≤M for all x∈[0, 2].

f''(x)=6x-6鈮�6x+6鈮�6x+6鈥�(1)

The expression (1)is the smallest when xis the smallest, which occurs when x=0for allx鈭�[0,2].

f''(x)鈮�6(0)+6鈮�6

The value of Mis 6for all values x鈭�[0,2].

04

Part (b) Step 1: Determine the value of M such that f''(x)≤M for all x∈[0, 4].

Consider the functionf(x)=x2(x-3).

Thus, the expression (1)is the smallest xwhen is the smallest, which occurs when x=0for all x鈭�[0,4].

The second

f''(x)鈮�6(0)+6鈮�6

Hence, the value of Mis 6for all values x鈭�[0,4].

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