University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 4 - Section 4.1 - Extreme Values of Functions - Exercises - Page 216: 60

Answer

Function $y$, at $x=-1$ and $x=3$, has the absolute and local minimum value of $0$. It also has the absolute and local maximum value of $2$ at $x=1$.
1543928367

Work Step by Step

$$y=\sqrt{3+2x-x^2}=(3+2x-x^2)^{1/2}$$ $y$ is not defined where $(3+2x-x^2)\ge0$. 1) Find all the critical points: Find the derivative of the function: $$y'=\frac{1}{2}(3+2x-x^2)^{-1/2}(3+2x-x^2)'$$ $$y'=\frac{1}{2}(3+2x-x^2)^{-1/2}(2-2x)$$ $$y'=\frac{2-2x}{2\sqrt{3+2x-x^2}}$$ $$y'=\frac{1-x}{\sqrt{3+2x-x^2}}$$ - For $y'=0$, $1-x=0$ or $x=1$. - For $y'$ to be undefined, $\sqrt{3+2x-x^2}=0$, or $(3+2x-x^2)=0$. We have $$3+2x-x^2=-(x^2-2x-3)=-\Big((x^2-2x+1)-4\Big)$$ $$=-\Big((x-1)^2-4\Big)$$ So $(3+2x-x^2)=0$ when $(x-1)^2=4$, or $(x-1)=\pm2$ For $(x-1)=2$, then $x=3$ and for $(x-1)=-2$, then $x=-1$. All these points $x=\{-1,1,3\}$ are critical points here. 2) Evaluate function $y$ at the critical points: - For $x=-1$: $$y=\sqrt{3+2\times(-1)-(-1)^2}=\sqrt{3-2-1}=0$$ - For $x=1$: $$y=\sqrt{3+2\times1-(1^2)}=\sqrt{3+2-1}=\sqrt4=2$$ - For $x=3$: $$y=\sqrt{3+2\times3-3^2}=\sqrt{3+6-9}=0$$ Now check the graph of function $y$. We see that function $y$, at $x=-1$ and $x=3$, has the absolute and local minimum value of $0$. It also has the absolute and local maximum value of $2$ at $x=1$.
Small 1543928367
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.