4.3 Consequences of differentiability, the mean value theorem  (Page 12/3)

Let $f:S\to R$ be a real-valued function of a real variable, and let $c$ be an element of the interior of $S.$ Then $f$ is said to attain a local maximum at $c$ if there exists a $\delta >0$ such that $(c-\delta ,c+\delta )\subseteq S$ and $f\left(c\right)\ge f\left(x\right)$ for all $x\in (c-\delta ,c+\delta ).$

The function $f$ is said to attain a local minimum at $c$ if there exists an interval $(c-\delta ,c+\delta )\subseteq S$ such that $f\left(c\right)\le f\left(x\right)$ for all $x\in (c-\delta ,c+\delta ).$

The next theorem should be a familiar result from calculus.

First derivative test for extreme values

Let $f:S\to R$ be a real-valued function of a real variable, and let $c\in S$ be a point at which $f$ attains a local maximum or a local minimum. If $f$ is differentiable at $c,$ then $f^{\text{'}}\left(c\right)$ must be $0.$

We prove the theorem when $f$ attains a local maximum at $c.$ The proof for the case when $f$ attains a local minimum is completely analogous.

Thus, let $\delta >0$ be such that $f\left(c\right)\ge f\left(x\right)$ for all $x$ such that $|x-c|<\delta .$ Note that, if $n$ is sufficiently large, then both $c+\frac{1}{n}$ and $c-\frac{1}{n}$ belong to the interval $(c-\delta ,c+\delta ).$ We evaluate $f^{\text{'}}\left(c\right)$ in two ways. First,

because the numerator is always nonpositive and the denominator is always positive.On the other hand,

since both numerator and denominator are nonpositive. Therefore, $f^{\text{'}}\left(c\right)$ must be 0, as desired.

Of course we do not need a result like [link] for functions of a complex variable, since the derivative of every real-valued function of a complex variable necessarily is 0,independent of whether or not the function attains an extreme value.

REMARK As mentioned earlier, the zeroes of a function are often important numbers. The preceding theorem shows that the zeroes of the derivative $f^{\text{'}}$ of a function $f$ are intimately related to finding the extreme values of the function $f.$ The zeroes of $f^{\text{'}}$ are often called the critical points for $f.$ Part (a) of the [link] establishes the familiar procedure from calculus for determining the maximum and minimum of a continuous real-valued function on a closed interval.

Probably the most powerful theorem about differentiation is the next one. It is stated as an equation, but its power is usually as an inequality; i.e., theabsolute value of the left hand side is less than or equal to the absolute value of the right hand side.

Mean value theorem

Let $f$ be a real-valued continuous function on a closed bounded interval $[a,b],$ and assume that $f$ is differentiable at each point $x$ in the open interval $(a,b).$ Then there exists a point $c\in (a,b)$ such that