One variable mean value and Taylor proofs

The segment arguments in several variables depend on one-variable theorems. Here those foundations are proved within MATH2066 rather than counted from another course. The difference between differentiability, continuous differentiability and interval-wide hypotheses remains visible throughout.

1 Derivatives and interior extrema

For $f$ defined on a neighborhood of $a$, $f'(a)$ is the finite limit $[f(a+h)-f(a)]/h$ as nonzero $h\to0$. It gives $f(a+h)=f(a)+f'(a)h+o(|h|)$ and therefore continuity. Sum and product rules follow by expanding increments; the chain rule follows by substituting local linear approximations. These rules require the relevant derivatives and, for a quotient, a nonzero denominator.

Fermat's necessary condition. At a differentiable interior local maximum or minimum, $f'(a)=0$.

Proof. At a local maximum, sufficiently small positive increments give nonpositive difference quotients and negative increments give nonnegative ones. A two-sided derivative can satisfy both sign limits only by being zero. Negating the function treats a minimum. $\square$

The interior hypothesis matters: $f(x)=x$ on $[0,1]$ has extrema at endpoints despite derivative 1. The converse fails: $x^3$ has zero derivative at zero but no extremum there.

For the product rule, put $\Delta f=f(a+h)-f(a)=f'(a)h+o(|h|)$ and similarly for $g$. Then

$$f(a+h)g(a+h)-f(a)g(a)=g(a)\Delta f+f(a)\Delta g+\Delta f\Delta g.$$

The last product is $O(h^2)$; division by $h$ gives $(fg)'(a)=f'(a)g(a)+f(a)g'(a)$. The sum rule follows by adding the two linear approximations. If $g(a)\ne0$, continuity keeps its nearby values nonzero and

$$\frac{1/g(a+h)-1/g(a)}h=-\frac{[g(a+h)-g(a)]/h}{g(a+h)g(a)}\longrightarrow-\frac{g'(a)}{g(a)^2}.$$

Combining reciprocal and product rules proves the quotient rule on its nonzero-denominator domain. For composition, the first increment is $k=f'(a)h+o(|h|)=O(|h|)$; substituting it into $g(f(a)+k)=g(f(a))+g'(f(a))k+o(|k|)$ gives the chain rule. If $k=0$, that remainder is zero, so the argument never requires dividing by a vanishing intermediate increment.

2 Rolle and mean value

Rolle's theorem. If $a<b$, $f$ is continuous on $[a,b]$, differentiable on $(a,b)$ and $f(a)=f(b)$, then $f'(c)=0$ for some $c\in(a,b)$.

Proof. The extreme-value theorem supplies maximum and minimum. If the function is constant, every interior point works. Otherwise a value differs from the equal endpoint values, so either a maximum or a minimum occurs in the interior. Fermat's condition applies there. $\square$

Mean value theorem. Under the same continuity and differentiability hypotheses but without equal endpoint values, some $c\in(a,b)$ satisfies $f(b)-f(a)=f'(c)(b-a)$.

Subtract the secant line $f(a)+[f(b)-f(a)](x-a)/(b-a)$ from $f$ and apply Rolle. This proves the theorem without assuming $f'$ is continuous. Consequences include: derivative zero on an interval makes the function constant; nonnegative derivative makes it nondecreasing; and $|f'|\le M$ on an interval gives $|f(y)-f(x)|\le M|y-x|$ for points in it. An interval-wide derivative bound cannot be inferred from one derivative value.

The derivative can be discontinuous while the theorem remains applicable. For $f(0)=0$, $f(x)=x^2\sin(1/x)$ otherwise, $f'(0)=0$ but $f'(x)=2x\sin(1/x)-\cos(1/x)$ oscillates near zero. Differentiability is weaker than the $C^1$ condition used by some integral proofs.

3 Cauchy's two-function version

For continuous $f,g$ on $[a,b]$, differentiable inside, Rolle applied to

$$H(x)=[f(b)-f(a)][g(x)-g(a)]-[g(b)-g(a)][f(x)-f(a)]$$

gives an interior $c$ with

$$[f(b)-f(a)]g'(c)=[g(b)-g(a)]f'(c).$$

If $g'$ never vanishes on $(a,b)$, Rolle also ensures $g(b)\ne g(a)$, so the equality can be divided to give a ratio formula. Without those nonzero conditions, the cross-multiplied equality is the correct statement. This distinction prevents an invalid division from being hidden inside a named theorem.

4 Taylor's theorem with a strict intermediate point

Suppose $f$ is $C^{n+1}$ on an open neighborhood of $[a,b]$, with $a<b$. There exists $\xi\in(a,b)$ such that

$$f(b)=\sum_{k=0}^n\frac{f^{(k)}(a)}{k!}(b-a)^k +\frac{f^{(n+1)}(\xi)}{(n+1)!}(b-a)^{n+1}.$$

Proof. Let $T_n$ be the displayed polynomial and choose $K$ so $H(t)=f(t)-T_n(t)-K(t-a)^{n+1}$ vanishes at $b$. Its value and first $n$ derivatives vanish at $a$. Rolle first gives a zero of $H'$ strictly between $a,b$; using $H'(a)=0$, apply Rolle again between $a$ and that point. Repeating gives a zero $\xi$ of $H^{(n+1)}$ strictly inside. There, $0=f^{(n+1)}(\xi)-(n+1)!K$, proving the formula. $\square$

For another target on the opposite side of the expansion point, use the same repeated-Rolle argument on the interval between them. If the target equals the expansion point, the remainder is zero; no strict intermediate point is asserted in that degenerate case.

A uniform bound $|f^{(n+1)}|\le M$ on the segment gives error at most $M|b-a|^{n+1}/(n+1)!$. The order depends on a neighborhood bound, not merely the derivative at $a$. The integral remainder proved elsewhere in this course gives a complementary form under its stated regularity.

5 One-variable convexity as a derivative statement

A differentiable function on an open interval is convex exactly when $f(y)\ge f(x)+f'(x)(y-x)$ for all points. Convexity gives this supporting-line inequality by dividing its defining chord inequality and taking a one-sided limit. Conversely, applying supporting inequalities at a point between $x,y$ and combining them with the chord weights cancels the derivative terms, yielding convexity.

If $f\in C^2$, convexity is equivalent to $f''\ge0$. One direction follows from the mean value theorem making $f'$ nondecreasing and integrating or comparing secant slopes. The other follows from nonnegative second differences and their derivative limit. A stationary point of a convex function is therefore a global minimum on its interval, whereas an arbitrary stationary point need not be.

Exercises with complete solutions

E1 Foundation. For $f=x^2$ on $[1,3]$, find the point supplied by the mean value theorem.

S1. The secant slope is $(9-1)/(3-1)=4$. Since $f'=2x$, the point is $c=2$, lying strictly between 1 and 3.

E2 Hypotheses. Why can Rolle not be applied to $|x|$ on $[-1,1]$ to infer a zero derivative inside?

S2. The function is not differentiable at zero, so an essential hypothesis fails. Its derivatives are $-1$ to the left and 1 to the right, with no zero derivative anywhere they exist. Equal endpoints and continuity are not enough.

E3 Taylor bound. Use the degree-2 Taylor polynomial of $e^x$ at zero to bound its error at $x=1$, and state the remainder at $x=0$ separately.

S3. The polynomial is $1+x+x^2/2$. On $[0,1]$, the third derivative is $e^x\le e$, so error at 1 is at most $e/6$. Taylor gives a strict intermediate $\xi\in(0,1)$ for that nonzero step. At $x=0$ the approximation is exact with remainder zero, not a claim about an intermediate point in an empty interval.

E4 Derivative discontinuity. Verify the derivative at zero of $x^2\sin(1/x)$ extended by zero, and exhibit two sequences showing its derivative is not continuous there.

S4. The difference quotient at zero is $x\sin(1/x)\to0$. At $x_n=1/(2\pi n)$, the derivative is $-1$; at $y_n=1/((2n+1)\pi)$, it is 1. Both sequences tend to zero but their derivative values do not approach $f'(0)=0$. The function remains differentiable, so MVT still applies on proper compact intervals.

Summary

Rolle follows from compact extrema and the derivative definition; the mean value theorem follows by subtracting a secant. Ratio versions require additional nonzero hypotheses. Repeated Rolle supplies a strict Taylor intermediate point only for a nonzero step, while convexity turns local derivative information into a global supporting inequality.