Question

Find \(f\)

\(f''\left( x \right) = \frac{2}{3}{x^{{2 \mathord{\left/

{\vphantom {2 3}} \right.

\kern-\nulldelimiterspace} 3}}}\)

Short answer

Required function is \(f\left( x \right) = \frac{3}{{20}}{x^{{8 \mathord{\left/

{\vphantom {8 3}} \right.

\kern-\nulldelimiterspace} 3}}} + Cx + D\).

Step-by-step solution

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

Here we have,

\(f''\left( x \right) = \frac{2}{3}{x^{{2 \mathord{\left/

{\vphantom {2 3}} \right.

\kern-\nulldelimiterspace} 3}}}\)

Now, the general antiderivative of \(f''\left( x \right) = \frac{2}{3}{x^{{2 \mathord{\left/

{\vphantom {2 3}} \right.

\kern-\nulldelimiterspace} 3}}}\)is,

\(\begin{aligned}{c}f'\left( x \right) = \frac{2}{3}\frac{{{x^{{5 \mathord{\left/

{\vphantom {5 3}} \right.

\kern-\nulldelimiterspace} 3}}}}}{{{\raise0.7ex\hbox{$5$} \!\mathord{\left/

{\vphantom {5 3}}\right.\kern-\nulldelimiterspace}

\!\lower0.7ex\hbox{$3$}}}} + C\\ = \frac{2}{5}{x^{{5 \mathord{\left/

{\vphantom {5 3}} \right.

\kern-\nulldelimiterspace} 3}}} + C\end{aligned}\)

Find a general antiderivative of \(f'\)

Now, we have, \(f'\left( x \right) = \frac{2}{5}{x^{{5 \mathord{\left/

{\vphantom {5 3}} \right.

\kern-\nulldelimiterspace} 3}}} + C\)

Now, the general antiderivative of \(f'\left( x \right) = \frac{2}{5}{x^{{5 \mathord{\left/

{\vphantom {5 3}} \right.

\kern-\nulldelimiterspace} 3}}} + C\) is,

\(\begin{aligned}{c}f\left( x \right) &= \frac{2}{5}\frac{{{x^{{8 \mathord{\left/

{\vphantom {8 3}} \right.

\kern-\nulldelimiterspace} 3}}}}}{{{\raise0.7ex\hbox{$8$} \!\mathord{\left/

{\vphantom {8 3}}\right.\kern-\nulldelimiterspace}

\!\lower0.7ex\hbox{$3$}}}} + Cx + D\\ = \frac{3}{{20}}{x^{{8 \mathord{\left/

{\vphantom {8 3}} \right.

\kern-\nulldelimiterspace} 3}}} + Cx + D\end{aligned}\)

So, our required function is\(f\left( x \right) = \frac{3}{{20}}{x^{{8 \mathord{\left/

{\vphantom {8 3}} \right.

\kern-\nulldelimiterspace} 3}}} + Cx + D\).