Answered: Find the x-value for all relative…

Functions and Change: A Modeling Approach to College Algebra (MindTap Course List)

6th Edition

ISBN:9781337111348

Author:Bruce Crauder, Benny Evans, Alan Noell

Publisher:Bruce Crauder, Benny Evans, Alan Noell

Chapter1: Functions

Section1.2: Functions Given By Tables

Problem 32SBE: Does a Limiting Value Occur? A rocket ship is flying away from Earth at a constant velocity, and it...

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Question

Find the x-value for all relative extrema of f(x) using the first derivative test. f(x) = 1 + x + x2 / x2 + x − 2 (this is a fraction)
Specify whether the x-value corresponds to a relative maximum or a relative minimum.

Expert Solution

Step 1

Given:

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

By first derivative test the critical point is at,

$f' (x) = 0$

Differentiate the function 'f(x)' with respect to x.

$\begin{array}{rcl} f' (x) & = & \frac{d}{d x} f (x) \\ = & \frac{d}{d x} (\frac{1 + x + x^{2}}{x^{2} + x - 2}) \\ = & \frac{(x^{2} + x - 2) \times \frac{d}{d x} (1 + x + x^{2}) - (1 + x + x^{2}) \times \frac{d}{d x} (x^{2} + x - 2)}{{(x^{2} + x - 2)}^{2}} \\ = & \frac{(x^{2} + x - 2) \times (1 + 2 x) - (1 + x + x^{2}) \times \frac{d}{d x} (2 x + 1)}{{(x^{2} + x - 2)}^{2}} \\ = & \frac{(1 + 2 x) [(x^{2} + x - 2) - (1 + x + x^{2})]}{{(x^{2} + x - 2)}^{2}} \\ = & \frac{(1 + 2 x) \times (- 3)}{{(x^{2} + x - 2)}^{2}} \\ = & \frac{- 3 (1 + 2 x)}{{(x^{2} + x - 2)}^{2}} \end{array}$

Step 2

Equate the above expression of $\begin{array}{rcl} f \end{array} \begin{array}{rcl} ' \end{array} \begin{array}{rcl} (x) \end{array}$ to zero and find the potential critical points.

$\begin{array}{rcl} \frac{d}{d x} (f (x)) & = & 0 \\ \frac{- 3 (1 + 2 x)}{{(x^{2} + x - 2)}^{2}} & = & 0 \\ - 3 (1 + 2 x) & = & 0 \\ (1 + 2 x) & = & 0 \\ x & = & \frac{- 1}{2} \\ a n d \\ {(x^{2} + x - 2)}^{2} & \neq & 0 \\ (x^{2} + x - 2) & \neq & 0 \\ (x + 2) (x - 1) & \neq & 0 \\ x & \neq & 1, - 2 \end{array}$

Thus, x=-1/2 is the potential critical point or relative extrema of the function f(x).

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