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: 11



Work Step by Step

The question asks us to find the equation of the tangent of the function $f(x)=x^2 + 1$ at the point $(2,5)$ 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: $f'(x)=2x$ $f'(2)=4$ Thus, the gradient of that function at the value x=2 is 4. 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 $(2,5)$ and $(0,c)$ (where c is the y-intercept). $\frac{5-c}{2-0}=4$ Solving for c gives us $c=-3$ Thus, the equation of the tangent at $(2,5)$ is $y=4x-3$
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