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: 17



Work Step by Step

The question asks us to find the equation of the tangent of the function $f(x)=\sqrt x=x^{\frac{1}{2}}$ at the point (4,2) 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: $f(x)=\sqrt x=x^{\frac{1}{2}}$ $f′(x)=\frac{1}{2}x^{\frac{1}{2}-1}=\frac{1}{2}x^{\frac{-1}{2}}=\frac{1}{2\sqrt x}$ $f′(4)=\frac{1}{2\sqrt 4}=0.25$ Thus, the gradient of that function at the value x=4 is 0.25. 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 (4,2) and (0,c) (where c is the y-intercept). $\frac{2−c}{4-0}=\frac{1}{4}$ Solving for c gives us c=1. Thus, the equation of the tangent at (1,4) is $y=\frac{1}{4}x+1$
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