The graph is drawn in such a way to draw attention to certain points. It certainly seems that the smallest \(y\)-value is \(-27\text{,}\) found when \(x=3\text{.}\) It also seems that the largest \(y\)-value is \(25\text{,}\) found at the endpoint of \(I\text{,}\) \(x=5\text{.}\) We use the word seems, for by the graph alone we cannot be sure the smallest value is not less than \(-27\text{.}\) Since the problem asks for an approximation, we approximate the extreme values to be \(25\) and \(-27\text{.}\)

We still do not have the tools to exactly find the relative extrema, but the graph does allow us to make reasonable approximations. It seems \(f\) has relative minima at \(x=-1\) and \(x=2\text{,}\) with values of \(f(-1)=0\) and \(f(2) = -5.4\text{.}\) It also seems that \(f\) has a relative maximum at the point \((0,1)\text{.}\)

We approximate the relative minima to be \(0\) and \(-5.4\text{;}\) we approximate the relative maximum to be \(1\text{.}\)

It is straightforward to evaluate \(\fp(x) =\frac{1}{5}\left(12x^3-12x^2-24x\right)\) at \(x=0, 1\) and \(2\text{.}\) In each case, \(\fp(x) = 0\text{.}\)

The figure implies that \(f\) does not have any relative maxima, but has a relative minimum at \((1,2)\text{.}\) In fact, the graph suggests that not only is this point a relative minimum, \(y=f(1)=2\) the minimum value of the function.

We compute \(\fp(x) = \frac{2}{3\sqrt[3]{x-1}}\text{.}\) When \(x=1\text{,}\) \(\fp\) is undefined.

We follow the steps outlined in Key Idea 3.1.16. We first evaluate \(f\) at the endpoints: \begin{align*} f(0)\amp=0\amp f(3)\amp=45. \end{align*}

Next, we find the critical values of \(f\) on \([0,3]\text{.}\) \(\fp(x) = 6x^2+6x-12 = 6(x+2)(x-1)\text{;}\) therefore the critical values of \(f\) are \(x=-2\) and \(x=1\text{.}\) Since \(x=-2\) does not lie in the interval \([0,3]\text{,}\) we ignore it. Evaluating \(f\) at the only critical number in our interval gives: \(f(1) = -7\text{.}\)

Table 3.1.19 gives \(f\) evaluated at the “important” \(x\)-values in \([0,3]\text{.}\) We can easily see the maximum and minimum values of \(f\text{:}\) the maximum value is \(45\) and the minimum value is \(-7\text{.}\)

Here \(f\) is piecewise-defined, but we can still apply Key Idea 3.1.16. Evaluating \(f\) at the endpoints gives: \begin{align*} f(-4)\amp=25\amp f(2)\amp=3. \end{align*}

We now find the critical numbers of \(f\text{.}\) We have to define \(\fp\) in a piecewise manner; it is \begin{equation*} \fp(x) =\begin{cases}2(x-1) \amp x \lt 0 \\ 1 \amp x>0\end{cases}. \end{equation*}

Note that while \(f\) is defined for all of \([-4,2]\text{,}\) \(\fp\) is not, as the derivative of \(f\) does not exist when \(x=0\text{.}\) (From the left, the derivative approaches \(-2\text{;}\) from the right the derivative is \(1\text{.}\)) Thus one critical number of \(f\) is \(x=0\text{.}\)

We now set \(\fp(x) = 0\text{.}\) When \(x >0\text{,}\) \(\fp(x)\) is never \(0\text{.}\) When \(x\lt 0\text{,}\) \(\fp(x)\) is also never \(0\text{.}\) (We may be tempted to say that \(\fp(x) = 0\) when \(x=1\text{.}\) However, this is nonsensical, for we only consider \(\fp(x) = 2(x-1)\) when \(x\lt 0\text{,}\) so we will ignore a solution that says \(x=1\text{.}\))

So we have three important \(x\)-values to consider: \(x= -4, 2\) and \(0\text{.}\) Evaluating \(f\) at each gives, respectively, \(25\text{,}\) \(3\) and \(1\text{,}\) shown in Table 3.1.21. Thus the absolute minimum of \(f\) is 1; the absolute maximum of \(f\) is \(25\text{.}\) Our answer is confirmed by the graph of \(f\) in Figure 3.1.22.

We again use Key Idea 3.1.16. Evaluating \(f\) at the endpoints of the interval gives: \(f(-2) = f(2) = \cos(4) \approx -0.6536\text{.}\) We now find the critical values of \(f\text{.}\)

Applying the Chain Rule, we find \(\fp(x) = -2x\sin\mathopen{}\left(x^2\right)\mathclose{}\text{.}\) Set \(\fp(x) = 0\) and solve for \(x\) to find the critical values of \(f\text{.}\)

We have \(\fp(x) = 0\) when \(x = 0\) and when \(\sin\mathopen{}\left(x^2\right)\mathclose{}\text{.}\) In general, \(\sin(t) = 0\) when \(t = \ldots -2\pi, -\pi, 0, \pi, \ldots\) Thus \(\sin\mathopen{}\left(x^2\right)\mathclose{} = 0\) when \(x^2 = 0, \pi, 2\pi, \ldots\) (\(x^2\) is always nonnegative so we ignore \(-\pi\text{,}\) etc.) So \(\sin\mathopen{}\left(x^2\right)\mathclose{}=0\) when \(x= 0, \pm \sqrt{\pi}, \pm\sqrt{2\pi}, \ldots\text{.}\) The only values to fall in the given interval of \([-2,2]\) are \(-\sqrt{\pi}\) and \(\sqrt{\pi}\text{,}\) approximately \(\pm 1.77\text{.}\)

We again construct a table of important values in Figure 3.1.24. In this example we have five values to consider: \(x= 0, \pm 2, \pm\sqrt{\pi}\text{.}\) From the table it is clear that the maximum value of \(f\) on \([-2,2]\) is \(1\text{;}\) the minimum value is \(-1\text{.}\) The graph in Figure 3.1.25 confirms our results.

A closed interval is not given, so we find the extreme values of \(f\) on its domain. \(f\) is defined whenever \(1-x^2\geq 0\text{;}\) thus the domain of \(f\) is \([-1,1]\text{.}\) Evaluating \(f\) at either endpoint returns \(0\text{.}\)

Using the Chain Rule, we find \(\fp(x) = -x\big/\sqrt{1-x^2}\text{.}\) The critical points of \(f\) are found when \(\fp(x) = 0\) or when \(\fp\) is undefined. It is straightforward to find that \(\fp(x) = 0\) when \(x=0\text{,}\) and \(\fp\) is undefined when \(x=\pm 1\text{,}\) the endpoints of the interval (which are in the domain of \(f\text{.}\)) The table of important values is given in Figure 3.1.28. The maximum value is \(1\text{,}\) and the minimum value is \(0\text{.}\)