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: 12



Work Step by Step

The question asks us to find the equation of the tangent of the function $f(x)=x -2x^2$ at the point $(1,-1)$ 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: $f'(x)=1-4x$ $f'(1)=-3$ Thus, the gradient of that function at the value x=1 is -3. 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 $(1,-1)$ and $(0,c)$ (where c is the y-intercept). $\frac{-1-c}{1-0}=-3$ Solving for c gives us $c=2$ Thus, the equation of the tangent at $(1,-1)$ is $y=-3x+2$
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