Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 3: Derivatives - Section 3.1 - Tangents and the Derivative at a Point - Exercises 3.1 - Page 108: 14



Work Step by Step

The question asks us to find the equation of the tangent of the function $f(x)=\frac{8}{x^2}$ at the point (2,2) Given that we are looking for the equation of a tangent, the first reasonable step would be to find the gradient at the points given, which can be found using the derivative function. Using the power rule: $f(x)=\frac{8}{x^2}=8x^{-2}$ $f′(x)=(-2)8x^{-2-1}=-16x^{-3}=\frac{-16}{x^{3}}$ $f′(2)=−2$ Thus, the gradient of that function at the value x=2 is -2. Using this information, we can find the intercept value of the tangent line. The equation of the gradient is given by: $\frac{y_{1}−y_{0}}{x_{1}−x_{0}}=m$ We can now use the points (2,2) and (0,c) (where c is the y-intercept). $\frac{2−c}{2-0}=-2$ Solving for c gives us c=6 Thus, the equation of the tangent at (3,3) is $y=−2x+6$
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