Question

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

20. \(f\left( x \right) = \ln \left( {2 + \sin x} \right)\)

Short answer

The function \(f\) is concave upward on \(\left( {\frac{{7\pi }}{6},\frac{{11\pi }}{6}} \right)\) and concave downward on \(\left( {0,\frac{{7\pi }}{6}} \right)\) and \(\left( {\frac{{11\pi }}{6},2\pi } \right)\). The inflection point is \(\left( {\frac{{7\pi }}{6},\ln \frac{3}{2}} \right)\) and \(\left( {\frac{{11\pi }}{6},\ln \frac{3}{2}} \right)\).

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 be concave downward on that interval.

Step 2: Find the solutions of the function

Consider the function \(f\left( x \right) = \ln \left( {2 + \sin x} \right)\).

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

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

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

\(\begin{aligned}{c}1 + 2\sin x < 0\\\sin x < - \frac{1}{2}\\\frac{{7\pi }}{6} < x < \frac{{11\pi }}{6}\end{aligned}\)

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

\(\begin{aligned}{c}1 + 2\sin x > 0\\\sin x > - \frac{1}{2}\\0 < x < \frac{{7\pi }}{6}\;{\rm{or }}\frac{{11\pi }}{6} < x < 2\pi \end{aligned}\).

Thus, \(f\) is concave upward on \(\left( {\frac{{7\pi }}{6},\frac{{11\pi }}{6}} \right)\) and concave downwardon \(\left( {0,\frac{{7\pi }}{6}} \right)\) and \(\left( {\frac{{11\pi }}{6},2\pi } \right)\).

Step 3: Find the inflection point

Consider the function\(f\left( x \right) = \ln \left( {2 + \sin x} \right)\).

As the inflection pointis at \(\left( {\frac{{7\pi }}{6},f\left( {\frac{{7\pi }}{6}} \right)} \right)\).and \(\left( {\frac{{11\pi }}{6},f\left( {\frac{{11\pi }}{6}} \right)} \right)\) Find \(f\left( {\frac{{7\pi }}{6}} \right)\) and \(f\left( {\frac{{11\pi }}{6}} \right)\).

\(\begin{aligned}{c}f\left( {\frac{{7\pi }}{6}} \right) = \ln \left( {2 + \sin \left( {\frac{{7\pi }}{6}} \right)} \right)\\ = \ln \frac{3}{2}\end{aligned}\)

And

\(\begin{aligned}{c}f\left( {\frac{{11\pi }}{6}} \right) = \ln \left( {2 + \sin \left( {\frac{{11\pi }}{6}} \right)} \right)\\ = \ln \frac{3}{2}\end{aligned}\)

Thus, the inflection points are \(\left( {\frac{{7\pi }}{6},\ln \frac{3}{2}} \right)\) and \(\left( {\frac{{11\pi }}{6},\ln \frac{3}{2}} \right)\).