Extrema and the Mean Value Theorem

Optimization problems ask for the shape of a can that minimizes material, the branching angle that minimizes the work of pumping blood, the production level that maximizes profit. Every such problem reduces to locating the largest or smallest value of a function. Two theorems make the search finite: the Extreme Value Theorem guarantees that a continuous function on a closed interval attains its extremes, and Fermat's Theorem pins the interior candidates to points where the derivative vanishes. The Mean Value Theorem then converts a bound on the derivative into a bound on the function itself.

Absolute and local extrema

A function can be large in two different senses: larger than every other value it takes, or merely larger than its immediate neighbors. Both matter, and they carry separate names.

Absolute extrema are also called global extrema, and the maximum and minimum values together are the extreme values of $f$. A local extremum is a purely neighborhood property; an absolute extremum is a statement about the whole domain. An endpoint can be an absolute extremum without being a local one, because a local extremum requires an open interval on both sides.

Some functions have extreme values and some do not. The function $f(x)=x^2$ attains its absolute minimum $0$ at the origin but has no maximum; $f(x)=x^3$ has no extreme values at all, local or absolute, since it is strictly increasing. The following theorem states exactly when existence is guaranteed.

Both hypotheses are needed. Drop continuity and a function can approach a value it never attains; drop the closed interval, replacing $[a,b]$ with the open $(a,b)$, and the same failure occurs at the missing endpoints. The theorem is plausible from a picture but its proof requires the completeness of the real numbers, so Stewart states it without proof.¹ It is an existence theorem: it promises the extrema exist but says nothing about where they are.

Fermat's Theorem and critical numbers

Locating an interior extremum uses one observation about the tangent line. At the top of a smooth hill or the bottom of a smooth valley, the tangent is horizontal.

The proof reads the sign of the difference quotient from both sides. Suppose $f$ has a local maximum at $c$. For small $h$, $f(c+h)\le f(c)$, so $f(c+h)-f(c)\le 0$. Dividing by $h>0$ keeps the sign, and dividing by $h<0$ flips it:

Both one-sided limits equal $f^{\prime }(c)$ because the derivative exists, so $f^{\prime }(c)\le 0$ and $f^{\prime }(c)\ge 0$ at once, forcing $f^{\prime }(c)=0$.

Two cautions keep Fermat's Theorem from being misread. Its converse is false: $f^{\prime }(c)=0$ does not imply an extremum. And it says nothing when $f^{\prime }(c)$ fails to exist, yet an extremum can still occur there.

Both failure modes point to the same set of candidates: the places where the derivative is zero, plus the places where it does not exist.

Fermat's Theorem, restated: if $f$ has a local extremum at $c$, then $c$ is a critical number. Local extrema occur only at critical numbers, so the search narrows to that (usually finite) list.

For example, the critical numbers of $f(x)=x^{3/5}(4-x)$ come from the product rule. Writing $f(x)=4x^{3/5}-x^{8/5}$,

The numerator vanishes at $x=\frac{3}{2}$, giving $f^{\prime }(x)=0$; the denominator vanishes at $x=0$, where $f^{\prime }$ does not exist. Both $0$ and $\frac{3}{2}$ are critical numbers, so both must be tested.

The Closed Interval Method

On a closed interval an absolute extremum is either a local extremum, which lands on a critical number by Fermat's Theorem, or it occurs at an endpoint. That dichotomy is exhaustive, so a three-step procedure always finds the absolute extrema of a continuous function on $[a,b]$.

The method is finite because the candidate list is finite. There is no need to decide which critical numbers are maxima and which are minima; comparing the values sorts that out at the end.

Consider $f(x)=x^3-3x^2+1$ on $\left[-\frac{1}{2},4\right]$. Since $f$ is a polynomial it is continuous, and $f^{\prime }(x)=3x^2-6x=3x(x-2)$ exists everywhere, so the only critical numbers are $x=0$ and $x=2$, both inside the interval. Tabulating the four candidates:

The absolute maximum is $f(4)=17$ and the absolute minimum is $f(2)=-3$. The maximum happens to sit at an endpoint and the minimum at an interior critical number; the method does not distinguish, since it only compares the tabulated values.

Rolle's Theorem

The Mean Value Theorem follows from a special case in which the endpoints share a height. If a differentiable function returns to the same value, its derivative must vanish somewhere between.

The proof splits into three cases. If $f$ is constant, then $f^{\prime }=0$ everywhere and any $c$ works. Otherwise $f$ takes some value above or below the common endpoint height. Suppose it rises above: by the Extreme Value Theorem $f$ attains a maximum on $[a,b]$, and since the endpoints share the lower value that maximum occurs at an interior $c$. Then $f$ has a local maximum at $c$, $f$ is differentiable there, and Fermat's Theorem gives $f^{\prime }(c)=0$. The case where $f$ dips below is symmetric, using the minimum.

To show $x^3+x-1=0$ has exactly one real root, first note the Intermediate Value Theorem forces a root: $f(0)=-1<0$ and $f(1)=1>0$. Suppose there were two roots $a$ and $b$. Then $f(a)=f(b)=0$, and Rolle's Theorem would produce a $c$ with $f^{\prime }(c)=0$. But $f^{\prime }(x)=3x^2+1\ge 1$ is never zero, a contradiction. So the root is unique.

The Mean Value Theorem

The Mean Value Theorem generalizes Rolle's Theorem. Instead of requiring equal endpoint heights, it compares the derivative to the average slope across the interval.

The right-hand quotient is the slope of the secant line through $A(a,f(a))$ and $B(b,f(b))$. The theorem asserts that at some interior point the tangent line is parallel to that secant.²

The proof applies Rolle's Theorem to the gap between $f$ and its own secant line. Let

the vertical distance from the curve down to the chord. Then $h$ is continuous on $[a,b]$, differentiable on $(a,b)$, and $h(a)=h(b)=0$, so Rolle's Theorem gives a $c$ with $h^{\prime }(c)=0$. Differentiating,

so $h^{\prime }(c)=0$ restates the Mean Value Theorem's conclusion.

Interpreted kinematically: if $s=f(t)$ is position, then the secant slope is the average velocity and $f^{\prime }(c)$ is the instantaneous velocity at time $c$. A car that covers $180$ km in $2$ hours averages $90$ km/h, so its speedometer read exactly $90$ km/h at some instant. More generally, the instantaneous rate of change equals the average rate of change somewhere on the interval.

Consequences

The Mean Value Theorem converts information about $f^{\prime }$ into information about $f$. Two consequences follow at once.

For any two points $x_1<x_2$ in the interval, the Mean Value Theorem gives a $c$ with $f(x_2)-f(x_1)=f^{\prime }(c)(x_2-x_1)=0$, so $f(x_2)=f(x_1)$. All values agree, so $f$ is constant. The interval hypothesis is essential: $f(x)=x/|x|$ has zero derivative on its whole domain yet is not constant, because that domain, $\{x:x\ne 0\}$, is not an interval.

Applying the previous theorem to $F=f-g$ gives $F^{\prime }=f^{\prime }-g^{\prime }=0$, hence $F$ is constant. Two functions with matching derivatives differ by a vertical shift. This is the fact that makes antiderivatives a family of curves differing by a constant, and it is the basis of the Fundamental Theorem of Calculus.

The Mean Value Theorem also bounds a function from a bound on its slope. If $f(0)=-3$ and $f^{\prime }(x)\le 5$ for all $x$, how large can $f(2)$ be? On $[0,2]$ there is a $c$ with $f(2)-f(0)=f^{\prime }(c)(2-0)$, so $f(2)=-3+2f^{\prime }(c)\le -3+2(5)=7$. The largest possible value of $f(2)$ is $7$.