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: 15



Work Step by Step

The question asks us to find the equation of the tangent of the function $h(t)=t^3$ at the point (2,8) 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: $h(t)=t^3$ $f′(x)=3t^{3-1}=3t^2$ $f′(2)=3(2)^2=12$ Thus, the gradient of that function at the value x=2 is 12. 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 (2,8) and (0,c) (where c is the y-intercept). $\frac{8−c}{2-0}=12$ Solving for c gives us c=-16 Thus, the equation of the tangent at (2,8) is $y=12x-16$
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