Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 3 - Applications of Differentiation - 3.1 Maximum and Minimum Values - 3.1 Execises - Page 212: 49


Absolute maximum: $f(-2)=33$ Absolute minimum: $f(2)=-31$.

Work Step by Step

Using closed interval method, as $f$ in continuous over the given range, let's first calculate the value of $f$ at critical numbers followed by values of $f$ at endpoints of the given range. For critical numbers, $f'(x)=0$ or $f'(x)$ should not exist. Let's differentiate $f(x)=3x^4-4x^3-12x^2+1$ with respect to $x$ keeping $[-2,3]$ as interval. $$\frac{df}{dx}=12x^3-12x^2-24x$$ $\frac{df}{dx}$ exists on the given interval. Thus, $\frac{df}{dx}=0$, $$12x^3-12x^2-24x=0\implies x^3-x^2-2x=0$$ $$x(x^2-x-2)=0$$ Thus, $x=0$ or $x^2-x-2=0\implies x=-1,2$. Therefore, three critical numbers $x=-1,0,2$. Thus, $f(-1)=-4$, $f(0)=1$ & $f(2)=-31$. At endpoints, $f(-2)=33$ & $f(3)=28$. Thus, absolute maximum $=f(-2)=33$ and absolute minimum $=f(2)=-31$.
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