Calculus, 10th Edition (Anton)

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

Chapter 2 - The Derivative - Chapter 2 Review Exercises - Page 185: 39

Answer

$y-1 = -3(x-\frac{3\pi}{4})$

Work Step by Step

We first solve for $f'(x)$: $f'(x) = Mcos(x)-Nsin(x)$ Thus, we have: $f(\frac{\pi}{4}) = Msin(\frac{\pi}{4}) + Ncos(\frac{\pi}{4}) = 3$ $f'(\frac{\pi}{4}) = Mcos(\frac{\pi}{4}) - Nsin(\frac{\pi}{4}) = 1$ $M\frac{\sqrt{2}}{2} + N\frac{\sqrt{2}}{2} = 3$ $M\frac{\sqrt{2}}{2} - N\frac{\sqrt{2}}{2}= 1$ Thus, adding the two equations, we get: $\sqrt{2}M = 4$ $M = 2\sqrt{2}$ Subtracting the two equations, we get: $\sqrt{2}N = 2$ $N = \sqrt{2}$ Thus, $f(x) = 2\sqrt{2}sin(x) + \sqrt{2}cos(x)$ and $f'(x) = 2\sqrt{2}cos(x) - \sqrt{2}sin(x)$ Therefore, $f(\frac{3\pi}{4}) = 2\sqrt{2}sin(\frac{3\pi}{4}) +\sqrt{2}cos(\frac{3\pi}{4}) = 1$ $f'(\frac{3\pi}{4}) = 2\sqrt{2}cos(\frac{3\pi}{4}) - \sqrt{2}sin(\frac{3\pi}{4}) = -3$ The equation for the tangent line is then: $y-1 = -3(x-\frac{3\pi}{4})$
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