Question

(a) Find the intervals on which f is increasing or decreasing.

(b) Find the local maximum and minimum values of f.

(c) Find the intervals of concavity and the inflection points.

28. \(f\left( x \right) = {\bf{co}}{{\bf{s}}^{\bf{2}}}x - {\bf{2sin}}\,x\)

Short answer

(a) The function is increasing in the interval \(\left( {\frac{\pi }{2},\frac{{3\pi }}{2}} \right)\) and decreasing in the intervals \(\left( {0,\frac{\pi }{2}} \right)\) and \(\left( {\frac{{3\pi }}{2},2\pi } \right)\).

(b) The local minimum value of the function is \( - 2\) , and the local maximum value of the function is 2.

(c) The function is concave upward in the interval \(\left( {\frac{\pi }{6},\frac{{5\pi }}{6}} \right)\), concave upward in the intervals \(\left( {0,\frac{\pi }{6}} \right)\), \(\left( {\frac{{5\pi }}{6},\frac{{3\pi }}{2}} \right)\), and \(\left( {\frac{{3\pi }}{2},2\pi } \right)\), and the points of inflection are \(\left( {\frac{\pi }{6}, - \frac{1}{4}} \right)\) and \(\left( {\frac{{5\pi }}{6}, - \frac{1}{4}} \right)\).

Step-by-step solution

Find the answer for part (a)

Differentiate the function \(f\left( x \right) = {\cos ^2}x - 2\sin x\).

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

By the inequality \(f'\left( x \right) > 0\),

\(\begin{aligned}{c}1 + \sin x \ge 0\\\sin x = - 1\\x = \frac{{3\pi }}{2}\end{aligned}\)

So, if

\(\cos x < 0\)

Then for the interval \(\frac{\pi }{2} < x < \frac{{3\pi }}{2}\), \(f'\left( x \right) > 0\).

Similarly, if \(\cos x > 0\)

Then for the intervals \(0 < x < \frac{\pi }{2}\) and \(\frac{{3\pi }}{2} < x < 2\pi \), \(f'\left( x \right) < 0\).

Therefore, the function is increasing in the interval \(\left( {\frac{\pi }{2},\frac{{3\pi }}{2}} \right)\) and decreasing in the interval \(\left( {0,\frac{\pi }{2}} \right)\) and \(\left( {\frac{{3\pi }}{2},2\pi } \right)\).

Find the answer for part (b)

The function is decreasing till \(x = \frac{\pi }{2}\) and increasing till \(x = \frac{{3\pi }}{2}\) and then again decreasing. Therefore, \(f\left( x \right)\) has a local minimum value of f will occur at \(x = - \frac{1}{2}\).

So, \(x = \frac{\pi }{2}\) is a point of local minimum and \(x = \frac{{3\pi }}{2}\) is the point of local maximum.

The local minimum value of the function is given below:

\(\begin{aligned}{c}f\left( {\frac{\pi }{2}} \right) = {\cos ^2}\left( {\frac{\pi }{2}} \right) - 2\sin \left( {\frac{\pi }{2}} \right)\\ = - 2\end{aligned}\)

The local maximum value of the function is given below:

\(\begin{aligned}{c}f\left( {\frac{{3\pi }}{2}} \right) = {\cos ^2}\left( {\frac{{3\pi }}{2}} \right) - 2\sin \left( {\frac{{3\pi }}{2}} \right)\\ = 2\end{aligned}\)

Thus, the local minimum value of the function is \( - 2\) and the local maximum value of the function is 2.

 Step 3: Find the answer for part (c)

Differentiate the equation \(f'\left( x \right) = - 2\cos x\left( {\sin x + 1} \right)\).

\(\begin{aligned}{c}f''\left( x \right) = \frac{{\rm{d}}}{{{\rm{d}}x}}\left( { - 2\cos x\left( {\sin x + 1} \right)} \right)\\ = - 2\left( {\cos x\frac{{\rm{d}}}{{{\rm{d}}x}}\left( {\sin x + 1} \right) + \left( {\sin x + 1} \right)\frac{{\rm{d}}}{{{\rm{d}}x}}\left( {\cos x} \right)} \right)\\ = - 2\left( {\cos x\left( {\cos x} \right) + \left( {\sin x + 1} \right)\left( { - \sin x} \right)} \right)\\ = - 2\left( {{{\cos }^2}x - {{\sin }^2}x - \sin x} \right)\\ = - 2\left( {1 - {{\sin }^2}x - {{\sin }^2}x - \sin x} \right)\\ = - 2\left( {1 - 2{{\sin }^2}x - \sin x} \right)\\ = 4{\sin ^2}x + 2\sin x - 2\\ = 2\left( {2\sin x - 1} \right)\left( {\sin x + 1} \right)\end{aligned}\)

For the inequality \(f''\left( x \right) > 0\),

\(\begin{aligned}{c}\sin x > \frac{1}{2}\\\frac{\pi }{6} < x < \frac{{5\pi }}{6}\end{aligned}\)

And

\(f''\left( x \right) < 0\)

\(\begin{aligned}{l}\sin x < \frac{1}{2}\\0 < x < \frac{\pi }{6}\,\,{\rm{or}}\,\,\frac{{5\pi }}{6} < x < \frac{{3\pi }}{2}\,\,\,{\rm{or}}\,\,\frac{{3\pi }}{2} < x < 2\pi \;\end{aligned}\)

So, the function is concave upward in the interval \(\left( {\frac{\pi }{6},\frac{{5\pi }}{6}} \right)\), and concave upward in the intervals \(\left( {0,\frac{\pi }{6}} \right)\), \(\left( {\frac{{5\pi }}{6},\frac{{3\pi }}{2}} \right)\), and \(\left( {\frac{{3\pi }}{2},2\pi } \right)\).

Find the value of \(f\left( x \right)\) at \(x = \frac{\pi }{6}\).

\(\begin{aligned}{c}f\left( {\frac{\pi }{6}} \right) = {\cos ^2}\left( {\frac{\pi }{6}} \right) - 2\sin \left( {\frac{\pi }{6}} \right)\\ = \frac{3}{4} - 1\\ = - \frac{1}{4}\end{aligned}\)

Find the value of \(f\left( x \right)\) at \(x = \frac{{5\pi }}{6}\)

\(\begin{aligned}{c}f\left( {\frac{{5\pi }}{6}} \right) = {\cos ^2}\left( {\frac{{5\pi }}{6}} \right) - 2\sin \left( {\frac{{5\pi }}{6}} \right)\\ = \frac{3}{4} - 1\\ = - \frac{1}{4}\end{aligned}\)

Therefore, the function is concave upward in the interval \(\left( {\frac{\pi }{6},\frac{{5\pi }}{6}} \right)\), concave upward in the intervals \(\left( {0,\frac{\pi }{6}} \right)\), \(\left( {\frac{{5\pi }}{6},\frac{{3\pi }}{2}} \right)\), and \(\left( {\frac{{3\pi }}{2},2\pi } \right)\), and the points of inflection are \(\left( {\frac{\pi }{6}, - \frac{1}{4}} \right)\) and \(\left( {\frac{{5\pi }}{6}, - \frac{1}{4}} \right)\).