Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 2 - Derivatives - 2.1 Derivatives and Rates of Change - 2.1 Exercises - Page 114: 22


If the tangent line to $y=f(x)$ at $(4,3)$ passes through the point $(0,2)$, find $f(4)$ and $f'(4)$.

Work Step by Step

We are given two points on the tangent line to $f(x)$ at $x=4$. These two points are $(4,3)$ and $(0,2)$. Using this information we can write the equation of the tangent line using point-slope form. The slope of the line is $(3-2)/(4-0) = 1/4$. Thus, $y-2 = 1/4(x)$ is the equation of the tangent line. $f'(4)$ is equal to the slope of the tangent line, $1/4$. Thus, $f'(4) = 1/4$ To find $f(4)$, plug in $4$ into the tangent line equation. This will yield $f(4)$ because the y-value of the tangent line at $x=4$ is the same as the y-value of the original $f(x)$ function at $x=4$. Plugging in $x=4$, we get $y-2 = 1/4(4)$ Rearranging the terms, we get $y = 3$ Thus, $f(4) = 3$
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