Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 4 - Applications of the Derivative - 4.2 Extreme Values - Exercises - Page 181: 12

Answer

There are no critical points.

Work Step by Step

$f'(t)=4-\frac{1}{2\sqrt {t^{2}+1}}(2t)$ $=\frac{8\sqrt {t^{2}+1}-2t}{2\sqrt {t^{2}+1}}$ As the denominator is always not equal to 0 and $t^{2}+1$ inside the square root is always positive, $f'(t)$ exists for all $t$. $f'(t)=0\implies8\sqrt {t^{2}+1}-2t=0$ $\implies64(t^{2}+1)-4t^{2}=0$ Or, $60t^{2}=-64$ There are no solutions for this equation. The critical points are those $t$ for which$f'(t)=0$ or $f'(t)$ doesn't exist. As there are no such points, this function has no critical points.
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