Question

Find \(f\left( x \right)\).

\(f''\left( x \right) = 1 - 6x + 48{x^2}\), \(f\left( 0 \right) = 1,f'\left( 0 \right) = 2\)

Short answer

The required function is, \(f\left( x \right) = \frac{1}{2}{x^2} - {x^3} + 4{x^4} + 2x + 1\).

Step-by-step solution

Find anti-derivative

Given that \(f''\left( x \right) = 1 - 6x + 48{x^2}\).

The general anti-derivative of the function is

\(\begin{aligned}{l}f'\left( x \right) &= x - \frac{{6{x^2}}}{2} + \frac{{48{x^3}}}{3} + C\\f'\left( x \right) &= x - 3{x^2} + 16{x^3} + C\end{aligned}\)

Find the constant

The obtained function is, \(f'\left( x \right) = x - 3{x^2} + 16{x^3} + C\).

Given that \(f'\left( 0 \right) = 2\)

\(\begin{aligned}{c}0 - 3{\left( 0 \right)^2} + 16{\left( 0 \right)^3} + C &= 2\\C &= 2\end{aligned}\)

So \(f'\left( x \right) = x - 3{x^2} + 16{x^3} + 2\).

Take anti-derivative again

Taking anti-derivative again on both sides of \(f'\left( x \right) = x - 3{x^2} + 16{x^3} + 2\).

\(\begin{aligned}{l}f\left( x \right) &= \frac{{{x^2}}}{2} - \frac{{3{x^3}}}{3} + \frac{{16{x^4}}}{4} + 2x + C\\f\left( x \right) &= \frac{1}{2}{x^2} - {x^3} + 4{x^4} + 2x + C\end{aligned}\)

Now \(f\left( 0 \right) = 1\)

\(\begin{aligned}{c}\frac{1}{2}{\left( 0 \right)^2} - {0^3} + 4{\left( 0 \right)^4} + 2\left( 0 \right) + C &= 1\\C &= 1\end{aligned}\)

Hence \(f\left( x \right) = \frac{1}{2}{x^2} - {x^3} + 4{x^4} + 2x + 1\).