Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 2 - Derivatives - 2.1 Derivatives and Rates of Change - 2.1 Exercises: 10

Answer

(a) $-\frac{1}{2}a^{-3/2}$ (b) For point (1,1) $y-1=-\frac{1}{2}(x-1)$ For point (4,$\frac{1}{2}$) $y-\frac{1}{2}=-\frac{1}{16}(x-4)$ (c) red: $y=\frac{1}{\sqrt x}$ purple: $y=-\frac{1}{2}x+\frac{3}{2}$ green: $y=-\frac{1}{16}x+\frac{3}{4}$
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Work Step by Step

(a) $y=\frac{1}{\sqrt x}$ at the point where $x=a$ Step 1: The question is asking for the slope of the tangent line so we have to find the first derivative. We can find the first derivative by using the power rule. $y=\frac{1}{\sqrt x}$ $y=x^{-1/2}$ $y'=-\frac{1}{2}x^{-3/2}$ Step 2: Plug in a wherever you see an x in the equation. $-\frac{1}{2}a^{-3/2}$ (b) points (1,1) and (4,$\frac{1}{2}$) Step 1: Plug in the given x values into $y'=-\frac{1}{2}x^{-3/2}$ $y'(1)=-\frac{1}{2}(1)^{-3/2}$ $y'(1)=-\frac{1}{2}$ $y'(4)=-\frac{1}{2}(4)^{-3/2}$ $y'(4)=-\frac{1}{16}$ Step 2: Now that we have calculated the slope of the tangent lines, we can write the equation of the tangent line in point slope form. For point (1,1) $y-1=-\frac{1}{2}(x-1)$ For point (4,$\frac{1}{2}$) $y-\frac{1}{2}=-\frac{1}{16}(x-4)$ Step 3: Rewrite the equations into standard form. For point (1,1) $y=-\frac{1}{2}x+\frac{3}{2}$ For point (4,$\frac{1}{2}$) $y=-\frac{1}{16}x+\frac{3}{4}$ (c) Plug the original function and equations of the tangent lines into a graphing calculator or graph by hand. red: $y=\frac{1}{\sqrt x}$ purple: $y=-\frac{1}{2}x+\frac{3}{2}$ green: $y=-\frac{1}{16}x+\frac{3}{4}$
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