Answered: f(x) x? - on [-4, 4] x? – 1

College Algebra (MindTap Course List)

12th Edition

ISBN:9781305652231

Author:R. David Gustafson, Jeff Hughes

Publisher:R. David Gustafson, Jeff Hughes

Chapter2: Functions And Graphs

Section2.CT: Chapter Test

Problem 3CT: Find f(1) and f(2). f(x)=xx1

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Question

a. Locate the critical points of ƒ.
b. Use the First Derivative Test to locate the local maximum and minimum values.
c. Identify the absolute maximum and minimum values of the function
on the given interval (when they exist).

Expert Solution

Step 1

Given: $f (x) = \frac{x^{2}}{x^{2} - 1}$ on the interval $[- 4, 4]$

To find critical points, find x such that $f' (x) = 0$

$f (x) = \frac{x^{2}}{x^{2} - 1} f' (x) = \frac{(x^{2} - 1) 2 x - x^{2} (2 x)}{{(x^{2} - 1)}^{2}} f' (x) = - \frac{2 x}{{(x^{2} - 1)}^{2}}$

Equate $f' (x)$ equal to 0

$\begin{array}{rcl} - \frac{2 x}{{(x^{2} - 1)}^{2}} & = & 0 \\ - 2 x & = & 0 \\ x & = & 0 \end{array}$

Thus, the critical point is x=0.

Step 2

Now the intervals therefore will be [-4, 0] and [0, 4]

Assign a test value for each interval to determine if Increasing or Decreasing.

Interval	x	$f' (x)$	Remarks
[-4, 0]	-2	$\frac{4}{9}$	Increasing
[0, ]	$- \frac{1}{2}$	$\frac{16}{9}$	Increasing
[0, 4]	2	$- \frac{4}{9}$	Decreasing

The change from increasing to decreasing corresponds to local maximum and the change from decreasing to increasing corresponds to local minimum.

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