Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 3 - Applications of Differentiation - 3.1 Exercises: 13

Answer

Critical Number is $\frac{8}{3}$.

Work Step by Step

Begin by taking the derivative of g(x) $g(x)= t\sqrt {4-t} $, $t<3$ $g'(x)= (\sqrt {4-t} )(1) + (t)(\frac{1}{2}(4-t)^{-\frac{1}{2}})(-1) $ Simplify $g'(x)= (4-t)^{\frac{1}{2}} - \frac{t}{2\sqrt {4-t}}$ $g'(x)= (4-t)^{\frac{1}{2}}(1-\frac{t}{2(4-t)}) $ Now set the derivative equal to zero to find the critical numbers $0=(4-t)^{\frac{1}{2}}$ $t=4$, but since $t<3$ it cannot be a critical number $0=(1-\frac{t}{2(4-t)})$ Simplify $t=\frac{8}{3}$ Because $\frac{8}{3} $<3. It is a critical number
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