Question

Find \(f\)

\(f''\left( x \right) = 20{x^3} - 12{x^2} + 6x\)

Short answer

Required function is \(f\left( x \right) = {x^5} - {x^4} + {x^3} + Cx + D\).

Step-by-step solution

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

Here we have,

\(f''\left( x \right) = 20{x^3} - 12{x^2} + 6x\)

Now, the general antiderivative of \(f''\left( x \right) = 20{x^3} - 12{x^2} + 6x\) is,

\(\begin{aligned}{c}f'\left( x \right) &= 20\frac{{{x^4}}}{4} - 12\frac{{{x^3}}}{3} + 6\frac{{{x^2}}}{2} + C\\ &= 5{x^4} - 4{x^3} + 3{x^2} + C\end{aligned}\)

Find a general antiderivative of \(f'\)

Now, we have, \(f'\left( x \right) = 5{x^4} - 4{x^3} + 3{x^2} + C\)

Now, the general antiderivative of \(f'\left( x \right) = 5{x^4} - 4{x^3} + 3{x^2} + C\) is,

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

So, our required function is \(f\left( x \right) = {x^5} - {x^4} + {x^3} + Cx + D\).