Question

Find \(f\)

\(f'\left( x \right) = 1 + 3\sqrt x ,\;f\left( 4 \right) = 25\)

Short answer

Required function is \(f\left( x \right) = x + 2{x^{{3 \mathord{\left/

{\vphantom {3 2}} \right.

\kern-\nulldelimiterspace} 2}}} + 5\).

Step-by-step solution

Find a general antiderivative of \(f'\)

Here we have,

\(f'\left( x \right) = 1 + 3\sqrt x \)

Now, the general antiderivative of \(f'\left( x \right) = 1 + 3\sqrt x \) is,

\(\begin{aligned}{c}f\left( x \right) &= x + 3\frac{{{x^{{3 \mathord{\left/

{\vphantom {3 2}} \right.

\kern-\nulldelimiterspace} 2}}}}}{{{\raise0.7ex\hbox{$3$} \!\mathord{\left/

{\vphantom {3 2}}\right.\kern-\nulldelimiterspace}

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

{\vphantom {3 2}} \right.

\kern-\nulldelimiterspace} 2}}} + C\end{aligned}\)

Find the value of constant \(C\)

Now, we have, \(f\left( x \right) = x + 2{x^{{3 \mathord{\left/

{\vphantom {3 2}} \right.

\kern-\nulldelimiterspace} 2}}} + C\)

Also we have \(f\left( 4 \right) = 25\)

Now, by putting \(f\left( 4 \right) = 25\) in \(f\left( x \right) = x + 2{x^{{3 \mathord{\left/

{\vphantom {3 2}} \right.

\kern-\nulldelimiterspace} 2}}} + C\)we get,

\(\begin{aligned}{c}25 = 4 + 2{\left( 4 \right)^{{3 \mathord{\left/

{\vphantom {3 2}} \right.

\kern-\nulldelimiterspace} 2}}} + C\\21 = 2\left( 8 \right) + C\\C = 5\end{aligned}\)

So, required function is \(f\left( x \right) = x + 2{x^{{3 \mathord{\left/

{\vphantom {3 2}} \right.

\kern-\nulldelimiterspace} 2}}} + 5\).