Question

17-22:Find the intervals on which\(f\)is concave upward or concave downward, and find the inflection points of\(f\).

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

Short answer

The function \(f\) is concave upward on \(\left( {0,\frac{\pi }{4}} \right)\) and \(\left( {\frac{{3\pi }}{4},\pi } \right)\) and concave downward on \(\left( {\frac{\pi }{4},\frac{{3\pi }}{4}} \right)\). The inflection point is \(\left( {\frac{\pi }{4},\frac{1}{2}} \right)\) and \(\left( {\frac{{3\pi }}{4},\frac{1}{2}} \right)\).

Step-by-step solution

Write the concavity test

If the function\(f''\left( x \right) > 0\)on an interval, then the graph of the function\(f\)is said to be concave upwardon that interval.

If the function \(f''\left( x \right) < 0\) on an interval, then the graph of the function \(f\) is said to beconcave downward on that interval.

Step 2: Find the solutions of the function

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

Differentiate the function w.r.t. \(x\) twice.

\(\begin{aligned}{c}f'\left( x \right) &= \frac{d}{{dx}}\left( {{{\sin }^2}x - \cos 2x} \right)\\ &= \frac{d}{{dx}}\left( {{{\sin }^2}x} \right) - \frac{d}{{dx}}\left( {\cos 2x} \right)\\ &= 2\sin x\cos x + 2\sin 2x\\ &= \sin 2x + 2\sin 2x\\ &= 3\sin 2x\\f''\left( x \right) &= \frac{d}{{dx}}\left( {3\sin 2x} \right)\\ &= 6\cos 2x\end{aligned}\)

If \(f''\left( x \right) > 0\) then we have,

\(\begin{aligned}{c}6\cos 2x &> 0\\\cos 2x &> 0\\0 < x &< \frac{\pi }{4}\;{\rm{and}}\;\frac{{3\pi }}{4} < x < \pi \end{aligned}\)

If \(f''\left( x \right) < 0\) then we have,

\(\begin{aligned}{c}\cos 2x &< 0\\\frac{\pi }{4} < x &< \frac{{3\pi }}{4}\end{aligned}\).

Thus, \(f\) isconcave upward on \(\left( {0,\frac{\pi }{4}} \right)\) and \(\left( {\frac{{3\pi }}{4},\pi } \right)\) and concave downward on \(\left( {\frac{\pi }{4},\frac{{3\pi }}{4}} \right)\).

Step 3: Find the inflection point

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

As the inflection pointis at \(\left( {\frac{\pi }{4},f\left( {\frac{\pi }{4}} \right)} \right)\).and \(\left( {\frac{{3\pi }}{4},f\left( {\frac{{3\pi }}{4}} \right)} \right)\) Find \(f\left( {\frac{\pi }{4}} \right)\) and \(f\left( {\frac{{3\pi }}{4}} \right)\).

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

And

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

Thus, theinflection points are \(\left( {\frac{\pi }{4},\frac{1}{2}} \right)\) and \(\left( {\frac{{3\pi }}{4},\frac{1}{2}} \right)\).