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.).

\(f\left( x \right) = \left\{ \begin{array}{l}{x^2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{if}} - 1 \le x \le 0\\2 - 3x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{if }}0 < x \le 1\end{array} \right.\)

Short answer

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

There is no local as well as an absolute maximum.

The local minimum is \(f\left( 0 \right) = 0\) , and the absolute minimum is \(f\left( 1 \right) = - 1\).

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:

\(f\left( x \right) = \left\{ \begin{array}{l}{x^2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{if}} - 1 \le x \le 0\\2 - 3x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{if }}0 < x \le 1\end{array} \right.\)

The graph is plotted as:

The absolute, as well as local maximum values, cannot be determined.

Now for the domain \(\left( { - 1,0} \right)\);

At \({\rm{x}} = 0\);

\(\begin{array}{c}f\left( x \right) = {0^2}\\ = 0\end{array}\)

This is the local minimum value.

Also, for the domain \(\left( {0,\left. 1 \right)} \right.\);

\(\begin{array}{c}f\left( x \right) = 2 - 3\left( 1 \right)\\ = - 1\end{array}\)

This is the absolute minimum value.

Hence, there is no local as well as an absolute maximum.

The local minimum is \(f\left( 0 \right) = 0\) , and the absolute minimum is \(f\left( 1 \right) = - 1\).