Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 3 - The Derivative In Graphing And Applications - 3.2 Analysis Of Functions II: Relative Extrema; Graphing Polynomials - Exercises Set 3.2 - Page 206: 41

Answer

Relative maxima $x=\frac{3}{2}$ Relative minima $x=3, x=0$

Work Step by Step

When $x \in(0,3) ; f(x)=3 x-x^{2} $ When $x \in(-\infty, 0) \cup(3, \infty) ; f(x)=-3 x+x^{2}$ Let $f_{1}(x)=-x^{2}+3 x, f_{2}(x)=x^{2}-3 x$ $f_{1}^{\prime}(x)=3-2 x; f_{2}^{\prime}(x)=-3+2 x$ So in $(0,3), 3 x-x^{2}=0$ at $x=\frac{3}{2}$ $\operatorname{In}(-\infty, 0) \cup(3, \infty), 2 x-3 \neq 0$ When we combine these results and generalize for all the values of x , we have that the stationary points of the function are $\frac{3}{2}=x$ The critical points also include the value of $x$ where the function is not differentiable. Hence, these values are $\mathrm{x}=0, \mathrm{x}=3$ Using the first derivative test, we know that: (a) If $f(x)>0$ on an open interval extending left from $x_{0}$ and $f(x)<0$ on an open interval extending right from $x_{0},$ then $f$ has a relative maximum at $x_{0}$ (b) If $f(x)<0$ on an open interval extending left from $x_{0}$ and $f(x)>0$ on an open interval extending right from $x_{0},$ then $f$ has a relative maximum at $x_{0}$ So by this test, $x=3$ and $x=0$ are values of minima.
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