University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Section 3.2 - The Derivative as a Function - Exercises: 13

Answer

$f'(x) =1-9x^{-2}$ $slope = 0$ Equation of tangent line: $y = -6$

Work Step by Step

$f(x) = x +\frac{9}{x}$ $f(-3) = -3 -\frac{9}{3} = -6$ Now we know the point on the graph is: (-3, -6). So, $f'(x) = 1+9(-1)x^{-1-1}$ $f'(x) =1-9x^{-2}$ Therefore, the slope at x = -3 is: $slope = 0$ So, the equation of tangent is: $\frac{y+6}{x+3} = 0$ y+6 = 0 y = -6
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