Question

Find \(f\)

\(f''\left( x \right) = {x^6} - 4{x^4} + x + 1\)

Short answer

Required function is \(f\left( x \right) = \frac{{{x^8}}}{{56}} - \frac{{2{x^6}}}{{15}} + \frac{{{x^3}}}{6} + Cx + D\).

Step-by-step solution

Find a general antiderivative of \(f''\)

Here we have,

\(f''\left( x \right) = {x^6} - 4{x^4} + x + 1\)

Now, the general antiderivative of \(f''\left( x \right) = {x^6} - 4{x^4} + x + 1\) is,

\(f'\left( x \right) = \frac{{{x^7}}}{7} - 4\frac{{{x^5}}}{5} + \frac{{{x^2}}}{2} + x + C\)

Find a general antiderivative of \(f'\)

Now, we have, \(f'\left( x \right) = \frac{{{x^7}}}{7} - 4\frac{{{x^5}}}{5} + \frac{{{x^2}}}{2} + x + C\)

Now, the general antiderivative of \(f'\left( x \right) = \frac{{{x^7}}}{7} - 4\frac{{{x^5}}}{5} + \frac{{{x^2}}}{2} + x + C\) is,

\(\begin{aligned}{c}f\left( x \right) &= \frac{{{x^8}}}{{7(8)}} - 4\frac{{{x^6}}}{{5(6)}} + \frac{{{x^3}}}{{2(3)}} + Cx + D\\ &= \frac{{{x^8}}}{{56}} - \frac{{2{x^6}}}{{15}} + \frac{{{x^3}}}{6} + Cx + D\end{aligned}\)

So, our required function is \(f\left( x \right) = \frac{{{x^8}}}{{56}} - \frac{{2{x^6}}}{{15}} + \frac{{{x^3}}}{6} + Cx + D\).