Question

1. Use a graph to estimate the absolute maximum and minimum values of the function to two decimal places.
2. Use calculus to find the exact maximum and minimum values.

72. \(f\left( x \right) = x - {\bf{2cos}}\,x\), \( - {\bf{2}} \le x \le {\bf{0}}\)

Short answer

1. The point \(f\left( { - 2} \right) = - 1.17\) is the abosolute maximum. The point \(f\left( { - 0.52} \right) = - 2.26\) is the absolute minimum.

2. The maximum value of \(f\left( x \right)\) is \( - 2 - 2\cos 2\) at \(x = - 2\) and the minimum value is \( - \frac{\pi }{6} - \sqrt 3 \) at \(x = - \frac{\pi }{6}\).

Step-by-step solution

(a) Step 1: Sketch the graph of \(f\left( x \right)\)

In the desmos graphing calculator (for closed interval) enter the expression \(x - 2\cos x\) in the tab and enter the interval \( - 2 \le x \le 0\), to plot the curve of \(f\left( x \right) = x - 2\cos x\).

The figure below represents the graph of \(f\left( x \right)\).

Find the point of maximum and minimum of \(f\left( x \right)\)

Following observations can be made from the graph of \(f\left( x \right)\).

1. The point \(f\left( { - 2} \right) = - 1.17\) is the abosolute maximum.
2. The point \(f\left( { - 0.52} \right) = - 2.26\) is the absolute minimum.

(b) Step 3: Differentiate the function \(f\left( x \right)\)

Find the derivative of \(f\left( x \right)\).

\(\begin{aligned}{c}f'\left( x \right) &= \frac{{\rm{d}}}{{{\rm{d}}x}}\left( {x - 2\cos x} \right)\\ &= 1 + 2\sin x\end{aligned}\)

Find the root of the equation \(f'\left( x \right) = 0\).

\(\begin{aligned}{c}1 + 2\sin x &= 0\\\sin x &= - \frac{1}{2}\\x &= - \frac{\pi }{6}\end{aligned}\)

Find the values of function at critical points of \(f\left( x \right)\)

The value of \(f\left( x \right)\) at \(x = - 2\);

\(\begin{aligned}{c}f\left( { - 2} \right) &= - 2 - 2\cos \left( { - 2} \right)\\ &= - 2 - 2\cos 2\end{aligned}\)

At \(x = - \frac{\pi }{6}\),

\(\begin{aligned}{c}f\left( { - \frac{\pi }{6}} \right) &= - \frac{\pi }{6} - 2\cos \left( { - \frac{\pi }{6}} \right)\\ &= - \frac{\pi }{6} - 2\left( {\frac{{\sqrt 3 }}{2}} \right)\\ &= - \frac{\pi }{6} - \sqrt 3 \end{aligned}\)

Thus, the maximum value of \(f\left( x \right)\) is \( - 2 - 2\cos 2\) at \(x = - 2\) and the minimum value is \( - \frac{\pi }{6} - \sqrt 3 \) at \(x = - \frac{\pi }{6}\).