Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 5 - Graphs and the Derivative - 5.1 Increasing and Decreasing Functions - 5.1 Exercises - Page 260: 23

Answer

$$ f(x)=\frac{x+2}{x+1} $$ (a) The critical numbers : There are no critical numbers. (b) No interval that the function is increasing , since $ f^{\prime}(x)$ is negative for all $x$ in the domain of $f$. (c) The function is decreasing on interval $(-\infty, -1)$ and $(-1, \infty)$.

Work Step by Step

$$ f(x)=\frac{x+2}{x+1} $$ Notice that the function $f$ is undefined when $x=-1$ , so $-1$ is not in the domain of $f$. To determine any critical numbers, first use the quotient rule to find $$ \begin{aligned} f^{\prime}(x) &=\frac{(x+1)(1)-(x+2)(1)}{(x+1)^{2}} \\ &=\frac{-1}{(x+1)^{2}} \end{aligned} $$ This derivative is never $0$, but it fails to exist at $x=-1$ , where the function is undefined. Since $-1$ is not in the domain of $f$, there are no critical numbers for $f$. We can still apply the first derivative test, however, to find where $f$ is increasing and decreasing. The number $-1$ (where $f $is undefined) divides the number line into two intervals: $(-\infty, -1)$ and $(-1, \infty)$. use a test point in each of these intervals to find that $f^{\prime}(x) \lt 0 $ for all $x$ except $-1$ . (This can also be determined by observing that $f^{\prime}(x) $ is the quotient of $-1$, which is negative, and $(x+1)^{2}$ , which is always positive or $0$.) This means that the function $f$ is decreasing on both $(-\infty, -1)$ and $(-1, \infty)$. Thus (a) The critical numbers : There are no critical numbers. (b) No interval that the function is increasing , since $ f^{\prime}(x)$ is negative for all $x$ in the domain of $f$. (c) The function is decreasing on interval $(-\infty, -1)$ and $(-1, \infty)$.
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