Finding Maximums and Minimums

One way to narrow our search for a maximum value of a function $f$ is to eliminate those values of $x$ which, for some reason, cannot possibly make $f$ maximum.

Proof: Assume that $\mathrm{f}^{\prime}(\mathrm{a})>0$. By definition, $\mathrm{f}^{\prime}(\mathrm{a})=\lim\limits _{\Delta x \rightarrow 0} \frac{f(a+\Delta x)-f(a)}{\Delta x}$, so $\lim \limits _{\Delta x \rightarrow 0} \frac{f(a+\Delta x)-f(a)}{\Delta x} > 0$ and the right and left limits are both positive: $\lim\limits _{\Delta x \rightarrow 0^{+}} \frac{f(a+\Delta x)-f(a)}{\Delta x} > 0$ and $\quad \lim\limits _{\Delta x \rightarrow 0^­} \frac{f(a+\Delta x)-f(a)}{\Delta x} > 0$.

Since the right limit, $\Delta x \rightarrow 0^{+}$, is positive, there are values of $\Delta x > 0$ so $\frac{f(a+\Delta x)-f(a)}{\Delta x} > 0$.

Multiplying each side of this last inequality by the positive $\Delta x$, we have $f(a+\Delta x)-f(a) > 0$ and $f(a+\Delta x)>f(a)$ so $f(a)$ is not a maximum.

Since the left limit, $\Delta x \rightarrow 0^­$, is positive, there are values of $\Delta x < 0$ so $\frac{f(a+\Delta x)-f(a)}{\Delta x} > 0$.

Multiplying each side of the last inequality by the negative $\Delta x$, we have that $f(a+\Delta x)-f(a) < 0$ and $f(a+\Delta x) < f(a)$ so $f(a)$ is not a minimum.'

The proof for the $"f '(a) < 0 "$ case is similar.

Example 1: Find the local extremes of $f(x)=x^{3}-6 x^{2}+9 x+2$ for all values of $x$.

Solution: An extreme value of $f$ can occur only where $f^{\prime}(x)=0$ or where $f$ is not differentiable. $f^{\prime}(x)=3 x^{2}-12 x+9=3\left(x^{2}-4 x+3\right)=3(x-1)(x-3)$ so $f^{\prime}(x)=0$ only at $x=1$ and $x=3 . f^{\prime}$ is a polynomial, so $\mathrm{f}$ is differentiable for all $\mathrm{x}$.

Practice 2: Find the local extremes of  $f(x)=x^{2}+4 x-5$  and  $g(x)=2 x^{3}-12 x^{2}+7$.

It is important to recognize that the conditions " $\mathrm{f}^{\prime}(\mathrm{a})=0$ " or " $\mathrm{f}$ not differentiable at a " do not guarantee that $\mathrm{f}(\mathrm{a})$ is a local maximum or minimum. They only say that $\mathrm{f}(\mathrm{a})$ might be a local extreme or that $\mathrm{f}(\mathrm{a})$ is a candidate for being a local extreme.

Example 2: Find all local extremes of $\mathrm{f}(\mathrm{x})=\mathrm{x}^{3}$.

Solution: $f(x)=x^{3}$ is differentiable for all $x$, and $f^{\prime}(x)=3 x^{2}$. The only place where

$f^{\prime}(x)=0$ is at $x=0$, so the only candidate is the point $(0,0)$. But if $x > 0$ then

$f(x)=x^{3} > 0=f(0)$, so $f(0)$ is not a local maximum. Similarly, if $x < 0$ then

$f(x)=x^{3} < 0=f(0)$ so $f(0)$ is not a local minimum. The point $(0,0)$ is the only candidate to be a local extreme of $\mathrm{f}$, and this candidate did not turn out to be a local extreme of $f$. The function $f(x)=x^{3}$ does not have any local extremes. (Fig. 5 )

Fig. 5

Practice 3: Sketch the graph of a differentiable function $\mathrm{f}$ which satisfies the conditions:
(i) $f(1)=5, f(3)=1, f(4)=3$ and $f(6)=7$,
(ii) $\mathrm{f}^{\prime}(1)=0, \mathrm{f}^{\prime}(3)=0, \mathrm{f}^{\prime}(4)=0$ and $\mathrm{f}^{\prime}(6)=0$,
(iii) the only local maximums of $f$ are at $(1,5)$ and $(6,7)$, and the only local minimum is at $(3,1)$.

Is f(a) a Maximum or Minimum or Neither?