Question

Find the derivative \(F\)of\(f\)that satisfies the given condition. Check your answer by comparing the graphs of\(f\)and\(F\).

\(f\left( x \right) = 4 - 3{\left( {1 + {x^2}} \right)^{ - 1}},\;F\left( 1 \right) = 0\)

Short answer

\(F\left( x \right) = 4x - 3{\tan ^{ - 1}}x + \frac{{3\pi }}{4} - 4\)

The graph is given by:

Step-by-step solution

Find the derivative of \(f\)

Here we have,

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

Let\(F\)be an antiderivative of\(f\).

To find the antiderivative we have to integrate the given function.

So,

\(\begin{aligned}{c}F\left( x \right) = \int {\left( {4 - \frac{3}{{1 + {x^2}}}} \right)dx} \\ = 4x - 3{\tan ^{ - 1}}x + C\end{aligned}\)

Given that, therefore we will substitute \(x = 1\) to find \(C\).

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

Substitute back the value of\(C\). We get,

\(F\left( x \right) = 4x - 3{\tan ^{ - 1}}x + \frac{{3\pi }}{4} - 4\)

Graph the function \(f\) and \(F\)

Graph of function \(f\left( x \right) = 4 - 3{\left( {1 + {x^2}} \right)^{ - 1}}\) and \(F\left( x \right) = 4x - 3{\tan ^{ - 1}}x + \frac{{3\pi }}{4} - 4\) is given by,

Now, we know that graph of \(f\) should be positive for \(\left( { - \infty ,\infty } \right)\) and the graph of \(F\) should be increasing, which is matched with given graph.