Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 4 - Section 4.1 - Maximum and Minimum Values - 4.1 Exercises: 36

Answer

$c=1-\sqrt5$ and $c=\sqrt5+1$ are critical numbers of $h$.

Work Step by Step

How to find the critical numbers of a function $f$ according to definition 1) Find all numbers of $c$ satisfying either $f'(c)=0$ or $f'(c)$ does not exist. 2) See that all the $c$ we get lie in the domain of $f$ $(D_f)$ or not. - If $c_0$ lies in $D_f$, $c_0$ is a critical number of $f$. - If $c_0$ does not lie in $D_f$, $c_0$ is not a critical number of $f$. $$h(p)=\frac{p-1}{p^2+4}$$ We see that equation $p^2+4\neq0$ for all values of $p$, for $p^2+4\gt0$ for all $p$, so the denominator here cannot be $0$. Therefore, $$D_h=R.$$ 1) Now, we find $h'(p)$ $$h'(p)=\frac{1(p^2+4)-(p-1)2p}{(p^2+4)^2}$$ $$h'(p)=\frac{p^2+4-2p^2+2p}{(p^2+4)^2}$$ $$h'(p)=\frac{-p^2+2p+4}{(p^2+4)^2}$$ We find $h'(c)=0$, $$h'(c)=0$$ $$-c^2+2c+4=0$$ $$c=1-\sqrt5\hspace{0.5cm}or\hspace{0.5cm}c=\sqrt5+1$$ There are no values of $c$ that makes $h'(c)$ not exist, for $(c^2+4)^2$ cannot equal $0$. 2) Examine whether $c$ lies in $D_h$ or not. We see that both $c=1-\sqrt5$ and $c=\sqrt5+1\in R$, so both lie in $D_h$. We conclude that $c=1-\sqrt5$ and $c=\sqrt5+1$ are critical numbers of $h$.
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