Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 3 - Derivatives - 3.3 Rules of Differentiation - 3.3 Exercises - Page 151: 38

Answer

The equation of the tangent line is $y=-\dfrac{3}{4}x+\dfrac{1}{4}$
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Work Step by Step

$y=\dfrac{e^{x}}{4}-x$ $;$ $a=0$ First, evaluate the derivative of the given expression: $y'=\dfrac{e^{x}}{4}-1$ Substitute $x$ by $a=0$ in the derivative found to obtain the slope of the tangent line at the given point: $m_{tan}=\dfrac{e^{0}}{4}-1=\dfrac{1}{4}-1=-\dfrac{3}{4}$ Substitute $x$ by $a=0$ in the original expression to obtain the $y$-coordinate of the point given: $y=\dfrac{e^{0}}{4}-0=\dfrac{1}{4}$ The point is $(0,\dfrac{1}{4})$ The slope of the tangent line and a point through which it passes are now known. Use the point-slope form of the equation of a line, which is $y-y_{1}=m(x-x_{1})$ to obtain the equation of the tangent line at the given point: $y-\dfrac{1}{4}=-\dfrac{3}{4}(x-0)$ $y-\dfrac{1}{4}=-\dfrac{3}{4}x$ $y=-\dfrac{3}{4}x+\dfrac{1}{4}$ The graph of both the function and the tangent line are shown in the answer section.
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