Calculus 8th Edition

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

Chapter 3 - Applications of Differentiation - 3.9 Antiderivatives - 3.9 Exercises - Page 283: 42

Answer

$$ f^{\prime \prime\prime}(x)=\cos x ,\quad f(0)=1 ,\quad f^{\prime}(0)=2, \quad f^{\prime\prime}(0)=3 $$ The required function is $$ f(x)=-\sin x+\frac{3}{2}x^{2}+3x+1 $$

Work Step by Step

$$ f^{\prime \prime\prime}(x)=\cos x ,\quad f(0)=1 ,\quad f^{\prime}(0)=2, \quad f^{\prime\prime}(0)=3 $$ The general anti-derivative of $ f^{\prime \prime\prime}(x)=\cos x $ is $$ f^{\prime \prime}(x)=\sin x+C $$ To determine C we use the fact that $f^{\prime\prime}(0)=3 $: $$ f^{\prime \prime}(0)=\sin (0)+C =3 $$ $ \Rightarrow $ $$ 0+C=3\quad \Rightarrow \quad C=3, $$ so $$ f^{\prime \prime}(x)=\sin x+3 $$ Using the anti-differentiation rules once more, we find that: $$ f^{\prime }(x)=-\cos x+3x+D $$ To determine D we use the fact that $f^{\prime}(0)=2 $: $$ f^{\prime }(0)=-\cos (0)+3(0)+D=2 $$ $ \Rightarrow $ $$ -1+D=2\quad \Rightarrow \quad D=3, $$ so $$ f^{\prime }(x)=-\cos x+3x+3 $$ Using the anti-differentiation rules once more, we find that: $$ f^(x)=-\sin x+\frac{3}{2}x^{2}+3x+E $$ To determine E we use the fact that $f(0)=1 $: $$ f(0)=-\sin (0)+\frac{3}{2}(0)^{2}+3(0)+E=1 $$ $ \Rightarrow $ $$ 0+E=1\quad \Rightarrow \quad E=1, $$ so the required function is $$ f(x)=-\sin x+\frac{3}{2}x^{2}+3x+1 $$
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.