Question

Find a function \(f\) such that \(f'\left( x \right) = {x^3}\) and the line \(x + y = 0\) is tangent to the graph of \(f\).

Short answer

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

Step-by-step solution

Find the point where given function passes through

Here we have, the line is\(x + y = 0 \Rightarrow y = - x\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left( 1 \right)\).

Now slope of line is -1.

\( \Rightarrow f'\left( x \right) = - 1\)

So,

\(\begin{aligned}{l}f'\left( x \right) &= - 1 &= {x^3}\\\therefore x &= - 1 \Rightarrow y &= 1\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left( {{\rm{by }}\left( 1 \right)} \right)\end{aligned}\)

So, the given function is pass through \(\left( { - 1,1} \right)\).

Find the required function

Here we have,\(f'\left( x \right) = {x^3}\)

Now, general antiderivative of\(f'\left( x \right) = {x^3}\)is,

\(f\left( x \right) = \frac{{{x^4}}}{4} + C\)

Also we have given function is pass through\(\left( { - 1,1} \right)\).

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

So, the required function is \(f\left( x \right) = \frac{{{x^4}}}{4} + \frac{3}{4}\).