Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 4 - Section 4.9 - Antiderivatives - 4.9 Exercises - Page 356: 50

Answer

$f(x)=\frac{1}{4}x^4+\frac{3}{4}$

Work Step by Step

In slope-intercept form, the equation of the tangent line is $y=-x$. Hence, the slope is $-1$. Since $f'(x)=x^3,$ we can plug in that slope to find the x-coordinate of the point at which the line is tangent to. $$-1=x^3 \Rightarrow x=-1.$$ Plugging that into the tangent line equation, we find that $y=1$. Now we just need to find the antiderivative of $x^3$ such that it hits the point $(-1,1)$. $$f(x)=\frac{1}{4}x^4+C$$ $$1=\frac{1}{4}\cdot{-1}^4+C \Rightarrow C=\frac{3}{4}$$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.