## Calculus 8th Edition

Published by Cengage

# Chapter 3 - Applications of Differentiation - 3.9 Antiderivatives - 3.9 Exercises - Page 283: 23

#### Answer

$$f(x) =x^{5}-x^{4}+x^{3}+C x+D$$ where $C,D$ are arbitrary constants .

#### Work Step by Step

$$f^{\prime \prime}(x)=20 x^{3}-12 x^{2}+6 x$$ The general anti-derivative of $f^{\prime \prime}(x)=20 x^{3}-12 x^{2}+6 x$ is $$\begin{split} f^{\prime}(x)& =20\left(\frac{x^{4}}{4}\right)-12\left(\frac{x^{3}}{3}\right)+6\left(\frac{x^{2}}{2}\right)+C \\ & =5 x^{4}-4 x^{3}+3 x^{2}+C \end{split}$$ Using the anti-differentiation rules once more, we find that $$\begin{split} f(x) &=5\left(\frac{x^{5}}{5}\right)-4\left(\frac{x^{4}}{4}\right)+3\left(\frac{x^{3}}{3}\right)+C x+D \\ &=x^{5}-x^{4}+x^{3}+C x+D \end{split}$$ where $C,D$ are arbitrary constants .

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