BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805
BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

Solutions

Chapter 4, Problem 3P
To determine

To examine: Whether the absolute maximum and absolute minimum of the function f(x)=e10|x2|x2 exist.

Expert Solution

Answer to Problem 3P

The absolute maximum of f(x)=e10|x2|x2 is f(5)=e45.

The absolute minimum of f(x)=e10|x2|x2 does not exist.

Explanation of Solution

The given function can be represented in the form f(x)=eg(x), where g(x)=10|x2|x2.

Thus, f(x) will have absolute maximum (minimum) when g(x) have absolute maximum (minimum).

Obtain the absolute maximum and minimum for g(x).

Express g(x) as follows.

g(x)={10(x2)x2ifx2>010[(x2)]x2ifx2<0

That is, g(x)={x2+10x20ifx>2x210x+20ifx<2

Obtain the first derivative of g(x).

g(x)={2x+10ifx>22x10ifx<2

Set g(x)=0 and obtain the critical numbers as follows.

2x+10=0x=52x10=0x=5

Since the derivative does not exist at x=2, it is also a critical point.

Therefore, the critical points are x=5,x=5andx=2.

Apply the critical points in f(x).

Substitute x=2 in f(x),

f(2)=e10|22|22=e4

Substitute x=5 in f(x),

f(5)=e10|52|52=e5

Substitute x=5 in f(x),

f(5)=e10|52|(5)2=e45

Since the largest functional value is the absolute maximum, the absolute maximum of f(x) is e45.

To find absolute minimum consider the following limit.

limxg(x)=limx(x2+10x20)=

Or

limxg(x)=limx(x210x+20)=

Therefore, limxg(x)=.

Using the above value of limit, find limxf(x).

limxf(x)=limxeg(x)=e=0

Since f(x)>0 as the exponential function is always positive for all values of x, so the absolute minimum of f(x) does not exist.

Therefore, the absolute maximum value of f(x) is f(5)=e45 and the absolute minimum of f(x) does not exist.

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