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.).

25. \(f\left( x \right) = 1 - \sqrt x \)

Short answer

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

There is no local maximum.

The absolute maximum is \(f\left( 0 \right) = 1\).

There is no local as well as an absolute minimum.

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:

\(f\left( x \right) = 1 - \sqrt x \)

Plot the function's graph \(f\left( x \right) = 1 - \sqrt x \) by reflecting it about the \(x\)-axis and then moving the curve 1 unit up in the positive \(y\)-axis.

The absolute maximum is as \(f\left( 0 \right) = 1\). But the local maximum cannot be found in this graph.

Thelocal, as well as absolute minimum, cannot be determined too.

At \(x = 0\),

\(\begin{array}{c}f\left( x \right) = 1 - \sqrt 0 \\ = 1\end{array}\)

This is the absolute maximum value.

Hence, there is no local maximum. The absolute maximum is \(f\left( 0 \right) = 1\).

There is no local as well as an absolute minimum for this graph.