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 - Page 283: 35


$c=0$ and $c=2$ are critical numbers of $g$.

Work Step by Step

How to find the critical numbers of a function according to definition 1) Find all numbers of $c$ satisfying whether $f'(c)=0$ 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 $c$ lies in $D_f$, $c$ is a critical number of $f$. - If $c$ does not lie in $D_f$, $c$ is not a critical number of $f$. $$g(y)=\frac{y-1}{y^2-y+1}$$ We see that $y^2-y+1\neq0$ for all values of $y$, so the denominator here cannot be $0$. Therefore, $$D_g=R$$ 1) Now, we find $g'(y)$ $$g'(y)=\frac{1(y^2-y+1)-(2y-1)(y-1)}{(y^2-y+1)^2}$$ $$g'(y)=\frac{y^2-y+1-2y^2+2y+y-1}{(y^2-y+1)^2}$$ $$g'(y)=\frac{-y^2+2y}{(y^2-y+1)^2}$$ - First, $$g'(c)=0$$ $$-c^2+2c=0$$ $$-c(c-2)=0$$ $$c=0\hspace{0.5cm}or\hspace{0.5cm}c=2$$ There are no values of $c$ that makes $g'(c)$ not exist, for $(c^2-c+1)^2$ cannot equal $0$. 2) Examine whether $c$ lies in $D_g$ or not. We see that both $c=0$ and $c=2\in R$, so both lie in $D_g$. We conclude that $c=0$ and $c=2$ are critical numbers of $g$.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.