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: 28



Work Step by Step

To find the equation of a tangent to the function $f(x)=\sqrt x$, it might be useful to first find the derivative function. $f(x)=\sqrt x=x^\frac{1}{2}$ $f'(x)=\frac{1}{2}x^{\frac{1}{2}-1}=\frac{1}{2}x^\frac{-1}{2}=\frac{1}{2\sqrt x}$ To find the x value at which the function has a slope of $\frac{1}{4}$, we equate the derivative function to $\frac{1}{4}$ and solve for $x$. $\frac{1}{4}=\frac{1}{2\sqrt x}$ Taking the reciprocal of both sides. $4=2\sqrt x$ $2=\sqrt x$ $x=4$ As such, it is the coordinate $(4,f(4))=(4,2)$ that has a slope of $\frac{1}{4}$. Solving for the y intercept of the tangent line; $\frac{2-c}{4-0}=\frac{1}{4}$ $c=1$ As such, the equation of the tangent is $y=\frac{1}{4}x+1$
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