Question

1. Graph the function.
2. Explain the shape of the graph by computing the limit as \(x \to {0^ + }\) or as \(x \to \infty \).
3. Estimate the maximum and minimum values and then use calculus to find the exact value.
4. Use a computer algebra system to compute \(f''\). Then use a graph of \(f''\) to estimate the \(x - \)coordinates of the inflection points.

26. \(f\left( x \right) = {\left( {\sin x} \right)^{\sin x}}\)

Short answer

1. The graph of the function is shown below:
2. 
   

1. The value of the limit is 1.
2. The estimate is confirmed by the approximations.
3. The graph is shown below:

The values of \(x - \)coordinates of inflection points are \(x \approx 0.94\) and \(x \approx 2.20\).

Step-by-step solution

Graph the function by a graphing calculator

a)

The procedure to draw the graph of the function by using the graphing calculator is as follows:

1. Open the graphing calculator. Select the “STAT PLOT” and enter the equation\(f\left( x \right) = {\left( {\sin x} \right)^{\sin x}}\)in the\({Y_1}\)tab.
2. Enter the “GRAPH” button in the graphing calculator.

Visualization of the graph of the function \(f\left( x \right) = {\left( {\sin x} \right)^{\sin x}}\) as shown below:

Explain the shape of the graph by computing the limit

b)

The function is \(y = f\left( x \right) = {\left( {\sin x} \right)^{\sin x}}\).

Take natural log on both sides of the equation as shown below:

\(\begin{array}{l}\ln y = \ln {\left( {\sin x} \right)^{\sin x}}\\\ln y = \sin x\ln \sin x\end{array}\)

Take the limit as \(x \to {0^ + }\) on both sides and evaluate the limit as shown below:

\(\begin{array}{c}\mathop {\lim }\limits_{x \to {0^ + }} \ln y = \mathop {\lim }\limits_{x \to {0^ + }} \sin x\ln \sin x\\ = \mathop {\lim }\limits_{x \to {0^ + }} \frac{{\ln \sin x}}{{\csc x}}\end{array}\)

It is observed that the limit is in indeterminate form \(\frac{\infty }{\infty }\).

Apply l’Hospital Rule as shown below:

\(\begin{array}{c}\mathop {\lim }\limits_{x \to {0^ + }} \frac{{\ln \sin x}}{{\csc x}} = \mathop {\lim }\limits_{x \to {0^ + }} \frac{{\frac{d}{{dx}}\left( {\ln \sin x} \right)}}{{\frac{d}{{dx}}\left( {\csc x} \right)}}\\ = \mathop {\lim }\limits_{x \to {0^ + }} \frac{{\cot x}}{{ - \csc x\cot x}}\\ = \mathop {\lim }\limits_{x \to {0^ + }} - \frac{1}{{\csc x}}\\ = \mathop {\lim }\limits_{x \to {0^ + }} \left( { - \sin x} \right)\\ = 0\\ \Rightarrow \mathop {\lim }\limits_{x \to {0^ + }} y\\ = {e^0}\\ = 1\end{array}\)

Thus, the value of the limit is 1.

Estimate the maximum and minimum values

It is observed that the local maximum value is at \(\left( {1.57,1} \right)\) and local minimum value at the point \(\left( {0.38,0.69} \right)\) and \(\left( {2.76,0.69} \right)\).

Take natural log on both sides of the equation as shown below:

\(\begin{array}{c}\ln y = \ln {\left( {\sin x} \right)^{\sin x}}\\\ln y = \sin x\ln \sin x\end{array}\)

Obtain the derivative as shown below:

\(\begin{array}{c}\frac{{y'}}{y} = \left( {\sin x} \right)\left( {\frac{{\cos x}}{{\sin x}}} \right) + \left( {\ln \sin x} \right)\cos x\\y' = y \cdot \cos x\left( {1 + \ln \sin x} \right)\\ = {\left( {\sin x} \right)^{\sin x}}\cos x\left( {1 + \ln \sin x} \right)\end{array}\)

Take \(y' = 0 \Rightarrow \cos x = 0\,\,\,{\mathop{\rm or}\nolimits} \,\,\,\ln \sin x = - 1\) to obtain as shown below:

\({x_2} = \frac{\pi }{2}\,\,\,{\mathop{\rm or}\nolimits} \,\,\,\sin x = {e^{ - 1}}\)

On the interval \(\left( {0,\pi } \right)\) to obtain as shown below:

\(\begin{array}{c}\sin x = {e^{ - 1}}\\{x_1} = {\sin ^{ - 1}}\left( {{e^{ - 1}}} \right)\end{array}\)

\({x_3} = \pi - {\sin ^{ - 1}}\left( {{e^{ - 1}}} \right)\)

Approximate the above points to obtain as shown below:

\(\left( {{x_1},f\left( {{x_1}} \right)} \right) \approx \left( {0.3767,0.6922} \right),\)\(\left( {{x_2},f\left( {{x_2}} \right)} \right) \approx \left( {1.5708,1} \right),\)and \(\left( {{x_3},f\left( {{x_3}} \right)} \right) \approx \left( {2.7649,0.6922} \right)\)

It is observed that estimates are confirmed by the approximations.

Use a computing device to compute \(f''\)

d)

The procedure to use a computer algebra system to solve the second derivative of the equation\(y = {\left( {\sin x} \right)^{\sin x}}\)as follows:

1. Open the derivative calculator. Enter the equation\({\left( {\sin x} \right)^{\sin x}}\)in the\(y\)tab.
2. Enter “2” in the number of times to differentiate tab.
3. Enter the “SUBMIT” button in the calculator.

Obtain the second derivative of the equation as shown below:

\(\begin{array}{l}f''\left( x \right) = \cos x{\sin ^{\left( {\sin x} \right)}}x\left( {\ln \left( {\sin x} \right) + 1} \right)\left( {\cos x\ln \left( {\sin x} \right) + \cos \left( x \right)} \right)\\ - {\sin ^{\sin \left( x \right) + 1}}\left( x \right)\left( {\ln \left( {\sin x} \right) + 1} \right) + {\cos ^2}\left( x \right){\sin ^{\sin x - 1}}x\end{array}\)

The procedure to draw the graph of the function by using the graphing calculator is as follows:

1. Open the graphing calculator. Select the “STAT PLOT” and enter the equation\(\begin{array}{l}f''\left( x \right) = \cos x{\sin ^{\left( {\sin x} \right)}}x\left( {\ln \left( {\sin x} \right) + 1} \right)\left( {\cos x\ln \left( {\sin x} \right) + \cos \left( x \right)} \right)\\ - {\sin ^{\sin \left( x \right) + 1}}\left( x \right)\left( {\ln \left( {\sin x} \right) + 1} \right) + {\cos ^2}\left( x \right){\sin ^{\sin x - 1}}x\end{array}\)in the\({Y_1}\)tab.
2. Enter the “GRAPH” button in the graphing calculator.

Visualization of the graph of the function as shown below:

It is observed from the graph that \(f''\left( x \right) = 0\) at \(x \approx 0.94\), and \(x \approx 2.20\). These values are \(x - \)coordinates of inflection points because \(f''\) varies signs at these values.