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: 75

Answer

$(-1,5)$ and $(1,1)$ and $(3,5)$ are the critical points of $y$. At $x=-1$ and $x=3$, function $y$ has the absolute and local maximum value of $5$, and at $x=1$, it has the local minimum value of $1$. There is no absolute minimum value.

Work Step by Step

$y=-x^2-2x+4$ for $x\le1$ and $y=-x^2+6x-4$ for $x\gt1$ - Domain: $R$. Therefore, there is no endpoint. 1) Find the derivative of function $y$: - For $x\le1$: $y'=-2x-2$ We have $y'=0$ when $-2x-2=0$, or $x=-1$. So there is a critical point at $x=-1$. - For $x\gt1$: $y'=-2x+6$ We have $y'=0$ when $-2x+6=0$, or $x=3$. So there is another critical point at $x=3$. - Also, we need to check whether $y'$ is defined at $x=1$ or not. To do so, let's check the graph of function $y$ below. The curve of $y$ at $x=1$ is not smooth enough for $y$ to be differentiable there. So $y'$ is undefined at $x=1$. That means there is one more critical point at $x=1$. 2) Evaluate $y$ at the critical points: - For $x=-1$, we would use the function $y=-x^2-2x+4$. $$y=-(-1)^2-2(-1)+4=-1+2+4=5$$ - For $x=1$, we would use the function $y=-x^2-2x+4$. $$y=-(1)^2-2\times1+4=-1-2+4=1$$ - For $x=3$, we would use the function $y=-x^2+6x-4$. $$y=-3^2+6\times3-4=-9+18-4=5$$ So $(-1,5)$ and $(1,1)$ and $(3,5)$ are the critical points of $y$. Now look at the graph of $y$. At $x=-1$ and $x=3$, function $y$ has the absolute and local maximum value of $5$, and at $x=1$, it has the local minimum value of $1$. There is no absolute minimum value.
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