MA005: Calculus I

The points at which a function might have an extreme value are called critical numbers.

If we are trying to find the extreme values of $\mathrm{f}$ on an open interval $\mathrm{c} < \mathrm{x} < \mathrm{d}$ or on the entire number line, then there will not be any endpoints so there will not be any endpoint critical numbers to worry about.

We can now give a very succinct description of where to look for extreme values of a function:

The critical numbers only give the possible locations of extremes, and some critical numbers are not the locations of extremes. The critical numbers are the candidates for the locations of maximums and minimums (Fig. 10). Section 3.5 is devoted entirely to translating and solving maximum and minimum problems.

Fig. 10

Which Functions Have Extremes?

So far we have concentrated on finding the extreme values of functions, but some functions don't have extreme values. Example 2 showed that $\mathrm{f}(\mathrm{x})=\mathrm{x}^{3}$ did not have a maximum or minimum.

Example 5: Find the extreme values of $\mathrm{f}(\mathrm{x})=\mathrm{x}$.

Solution: Since $\mathrm{f}^{\prime}(\mathrm{x})=1 > 0$ for all $\mathrm{x}$, the first theorem in this section guarantees that $\mathrm{f}$ has no extreme values. The function $\mathrm{f}(\mathrm{x})=\mathrm{x}$ does not have a maximum or minimum on the real number line.

The difficulty with the previous function was that the domain was so large that we could always make the function larger or smaller than any given value. The next example shows that we can encounter the same difficulty even on a small interval.

Example 6: Show that $\mathrm{f}(\mathrm{x})=\frac{1}{\mathrm{x}}$ (Fig. 11) does not have a maximum or minimum on the interval $(0,1)$.

Fig. 11

Solution: $\mathrm{f}$ is continuous for all $\mathrm{x} \neq 0$ so $\mathrm{f}$ is continuous on the interval $(0,1)$ For $0 < x < 1$, $f(x)=\frac{1}{x} > 0$ (Fig.11). For any number a strictly between $0$ and $1$, we can show that $\mathrm{f}(\mathrm{a})$ is neither a maximum nor a minimum of $\mathrm{f}$ on $(0,1)$. Pick $b$ to be any number between $0$ and $a$, $0 < b < a$. Then $\mathrm{f}(\mathrm{b})=\frac{1}{\mathrm{~b}} > \frac{1}{\mathrm{a}}=\mathrm{f}(\mathrm{a})$, so $\mathrm{f}(\mathrm{a})$ is not a maximum. Similarly, pick $\mathrm{c}$ to be any number between $a$ and $1$, $\mathrm{a} < \mathrm{c} < 1$. Then $\mathrm{f}(\mathrm{a})=\frac{1}{\mathrm{a}} > \frac{1}{\mathrm{c}} = \mathrm{f}(\mathrm{c})$, so $\mathrm{f}(\mathrm{a})$ is not a minimum. The interval $(0,1)$ is not large, but $\mathrm{f}$ still does not have an extreme value in $(0,1)$.

The Extreme Value Theorem gives conditions so that a function is guaranteed to have a maximum and a minimum.

The proof of this theorem is difficult and is omitted. Fig. 12 illustrates some of the possibilities for continuous and discontinuous functions on open and closed intervals. The Extreme Value Theorem guarantees that certain functions (continuous) on certain intervals (closed) must have maximums and minimums. Other functions on other intervals may or may not have maximums and minimums.

Fig. 12