Question

Find the absolute maximum and absolute minimum values of \(f\) on the given interval.

66. \(f\left( x \right) = x - 2{\tan ^{ - 1}}x,{\rm{ }}\left( {0,4} \right)\)

Short answer

The absolute maximum value of the function \(f\left( x \right) = x - 2{\tan ^{ - 1}}x\)is \(f\left( 4 \right) \approx 1.35\).

The absolute minimum value of the function \(f\left( x \right) = x - 2{\tan ^{ - 1}}x\)is \(f\left( 1 \right) \approx - 0.57\)

Step-by-step solution

The Closed Interval Method

There are three steps to find the absolute maximum and minimum values of a continuous function\(f\)in any closed interval\(\left( {a,b} \right)\):

1. Byfinding critical numbers, find the values of\(f\)at the critical numbers in \(\left( {a,b} \right)\).
2. Find the value of the function at eachendpoint of the interval.
3. Identify the largest value and smallest value obtained from step 1 and 2, wherelargest value will be absolute maximum value and the smallest value will be absolute minimum value.

Find the critical points

Consider the given function \(f\left( x \right) = x - 2{\tan ^{ - 1}}x\). Differentiate the function w.r.t \(x\) by using the formulas \(\frac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}\) and \(\frac{d}{{dx}}\left( {{{\tan }^{ - 1}}x} \right) = \frac{1}{{1 + {x^2}}}\).

\(\begin{aligned}{c}\frac{d}{{dx}}\left( {f\left( x \right)} \right) &= \frac{d}{{dx}}\left( {x - 2{{\tan }^{ - 1}}x} \right)\\f'\left( x \right) &= \frac{d}{{dx}}\left( x \right) - \frac{d}{{dx}}\left( {2{{\tan }^{ - 1}}x} \right)\\ &= 1 - 2\frac{d}{{dx}}\left( {{{\tan }^{ - 1}}x} \right)\\ &= 1 - 2 \cdot \frac{1}{{1 + {x^2}}}\end{aligned}\)

Substitute \(f'\left( x \right) = 0\) and solve for \(x\) to get the critical points.

\(\begin{aligned}{c}1 - 2 \cdot \frac{1}{{1 + {x^2}}} &= 0\\1 &= \frac{2}{{1 + {x^2}}}\\1 + {x^2} &= 2\\{x^2} &= 1\\x &= \pm 1\end{aligned}\)

Find value of \(f\)at critical points in \(\left( {0,4} \right)\)

At \(x = 1\),

\(\begin{aligned}{c}f\left( 1 \right) &= 1 - 2{\tan ^{ - 1}}1\\ &= 1 - \frac{\pi }{2}\\ &\approx - 0.57\end{aligned}\)

Do not find at \(x = - 1\), because it does not include in the interval \(\left( {0,4} \right)\).

Find value of \(f\) at endpoints of \(\left( {0,4} \right)\)

At \(x = 0\)

\(\begin{aligned}{c}f\left( 0 \right) &= 0 - 2{\tan ^{ - 1}}0\\ &= 0\end{aligned}\)

At \(x = 4\)

\(\begin{aligned}{c}f\left( 4 \right) &= 4 - 2{\tan ^{ - 1}}4\\ &\approx 1.35\end{aligned}\)

Observe the largest and smallest value of \(f\)

It can be observed that, at \(x = 4\), \(f\left( x \right) = 1.35\) which is the largest value that will be the absolute maximum value, and at \(x = 1\), \(f\left( x \right) = - 0.57\) or \(f\left( x \right) = 1 - \frac{\pi }{2}\) is the smallest value, that will be the absolute minimum value.

Thus, the absolute maximum and minimum values of the function\(f\left( x \right) = x - 2{\tan ^{ - 1}}x\) are \(f\left( 4 \right) \approx 1.35\) and \(f\left( 1 \right) \approx - 0.57\) respectively.