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: 16

Answer

$ θ=\frac{11\pi}{6} $, $ \frac{7\pi}{6} $

Work Step by Step

The first step is to take the derivative of the function $ f(θ)= 2sec(θ) + tan(θ) $. Using the Basic Differentiation Rules you get $ f'(θ)=2sec(θ)tan(θ) + (sec( θ ))^{2} $ Next Simplify $f'( θ )=2\frac{1}{cos( θ)} \times \frac{sin( θ )}{cos( θ )} + \frac{1}{(cos θ )^2} $ $f'( θ )= \frac{2sin(θ)}{cos^2(θ)} + \frac{1}{cos^2(θ)}$ $ f'(θ)= \frac{2sin(θ) +1}{cos^2(θ)} $ Then set the derivative equal to zero and solve for θ. $0=\frac{2sin(θ)+1}{cos^2(θ)}$ , Multiply both sides by $cos^2(θ)$ $0=2sin(θ)+1$, Isolate the $sin(θ) $ term $sin(θ)=-\frac{1}{2} $, take the $sin^{-1}(θ) $ of both sides to get $ θ=\frac{11\pi}{6} $, $ \frac{7\pi}{6} $
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