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Chapter 4: Q1P (page 213)

To find the familiar "second derivative test "for a maximum or minimum point. That is show that f'(a)=0, thenf''(a)>0implies a minimum point atx=aandf''(a)<0implies a maximum point at x=a.

Short Answer

Expert verified

For all xnear enough to ais given as follows:

f(x)-f(a)>0f(x)>f(a)

Step by step solution

01

Given data

The given function is f'(a)=0, and then f''(a)>0which implies a minimum point at x=a and f''(a)<0 implies a maximum point at x=a.

02

Concept of Taylor series

Taylor series representation off aboutx=ais given as:

f(x)=f(a)+f'(a)1!+f''(a)2!(x-a)2 鈥�(1)

03

Calculation to find the maximum point for the derivative test f

Suppose that f'(a)=0then from (1) obtain:

f''(a)>02f(x)-f(a)(x-a)2>0

For all xnear enough to ais given as follows:

f(x)-f(a)>0f(x)>f(a)

That is, fincreases.

Now, by the first derivative test fhas minimum point at x=a:

f''(a)<02f(x)-f(a)(x-a)<0f(x)-f(a)<0

f(x)<f(a)for all xnear enough to a.

That is, fdecreases.

Now, by the first derivative test fhas maximum point at x=a.

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