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.).

\(f\left( t \right) = \cos t,\,\,\,\,\,\, - \frac{{3\pi }}{2} \le t \le \frac{{3\pi }}{2}\)

Short answer

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

Step-by-step solution

Minima and Maxima of a function

TheCritical numbersfor 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:

\(f\left( t \right) = \cos t,\,\,\,\,\,\, - \frac{{3\pi }}{2} \le t \le \frac{{3\pi }}{2}\)

Here, \(f\left( t \right)\) is continuous in the interval \(\left( { - \frac{{3\pi }}{2},\frac{{3\pi }}{2}} \right)\). So, the graph is plotted as

The maximum and minimum values can be seen in the graph within the domain\(\left( { - \frac{{3\pi }}{2},\frac{{3\pi }}{2}} \right)\).

At\({\rm{t}} = 0\),

\(\begin{array}{c}f\left( t \right) = \cos 0^\circ \\ = 1\end{array}\)

This is absolute as well as the maximum local value.

Also,

At\({\rm{t}} = \pm \pi \),

\(\begin{array}{c}f\left( t \right) = \cos \left( { \pm \pi } \right)\\ = - 1\end{array}\)

This is absolute as well as the local minimum value.

Hence, the local maximum and the absolute maximum are\(f\left( 0 \right) = 1\).

The local minimum and the absolute minimum are \(f\left( { \pm \pi } \right) = - 1\).