Question

Sketch the graph of \(f\) by hand and use your sketch to find the absolute and local maximum and minimum values of \(f\). (Use the graphs and transformations of Sections 1.2 and 1.3.).

\(f\left( x \right) = \sin x,\,\,\,\,\,\, - \frac{\pi }{2} \le x \le \frac{\pi }{2}\)

Short answer

The graph of \(f\left( x \right)\)is:

There is no local maximum, and the absolute maximum is: \(f\left( {\frac{\pi }{2}} \right) = 1\).

There is no local minimum, and the absolute minimum is: \(f\left( { - \frac{\pi }{2}} \right) = - 1\).

Step-by-step solution

Minima and Maxima of a function

TheCritical numbers for any function\(f\left( x \right)\)can be obtained by putting\(f'\left( x \right) = 0\).

For the points \(a{\rm{ and }}b\)such that:

\(\begin{array}{l}f\left( a \right) \to {\rm{maximum}} \Rightarrow a\,\,{\rm{is maxima}}\\f\left( b \right) \to {\rm{minimum}}\,\,\, \Rightarrow a\,\,{\rm{is minima}}\end{array}\)

Graphing the function for minima and maxima:

The given function is:

\(f\left( x \right) = \sin x,\,\,\,\,\,\, - \frac{\pi }{2} \le x \le \frac{\pi }{2}\)

Here, \(f\left( x \right)\) is continuous in the interval \(\left( { - \frac{\pi }{2},\frac{\pi }{2}} \right)\). So, the graph is plotted as:

The maximum and minimum values can be seen in the graph within the domain \(\left( { - \frac{\pi }{2},\frac{\pi }{2}} \right)\). There are no local minima or maxima points found.

At \(x = \frac{\pi }{2}\),

\(\begin{array}{c}f\left( x \right) = \sin \frac{\pi }{2}\\ = 1\end{array}\)

This is the absolute maximum value.

Also,

At \(x = - \frac{\pi }{2}\),

\(\begin{array}{c}f\left( x \right) = \sin \left( { - \frac{\pi }{2}} \right)\\ = - 1\end{array}\)

This is the absolute minimum value.

Hence, there is no local maximum, and the absolute maximum is: \(f\left( {\frac{\pi }{2}} \right) = 1\).

There is no local minimum, and the absolute minimum is: \(f\left( { - \frac{\pi }{2}} \right) = - 1\).