Interior Extrema and the Vanishing Derivative

Fermat's theorem is the computational engine behind every "take the derivative and set it to zero" optimization technique taught across science and engineering. In physics, finding equilibrium points of a potential energy function $U(x)$ reduces to solving $U'(x)=0$, precisely because stable and unstable equilibria are local extrema of $U$. In economics, profit-maximizing quantities are found by setting marginal revenue equal to marginal cost — an instance of $f'(x)=0$ at an interior maximum of profit. In statistics and machine learning, maximum-likelihood estimation and loss-minimization in gradient-based training both begin from the same principle in many-variable form: at an interior optimum of a differentiable objective, every partial derivative vanishes, giving the gradient equation $\nabla f = 0$ used throughout regression, neural network training, and convex optimization. In numerical analysis, root-finding methods for $f'(x)=0$ (Newton's method applied to the derivative) are the standard computational route to locating extrema when no closed form exists. And within pure mathematics, Fermat's theorem is the base case from which Rolle's Theorem is proved (by applying Fermat's theorem to the interior point where a function returning to the same value attains its max or min), which in turn drives the Mean Value Theorem and the entire theory of monotonicity, convexity, and curve sketching.

:::quiz[Which hypothesis of Fermat's theorem fails for $f(x)=x$ on the domain $[0,1]$ at the point $c=0$, even though $f$ is a perfectly differentiable, simple function?]

• $f$ is not differentiable at $0$
• $c=0$ is not an interior point of the domain $[0,1]$ ✓
• $f$ does not have a local extremum at $0$
• The limit defining $f'(0)$ does not exist

Ask the tutor: Can you show me step by step how Fermat's theorem is used inside the proof of Rolle's Theorem?

Picture the graph of $f$ near an interior local maximum at $c$. Because $f$ is defined on both sides of $c$ and $f(c)$ is at least as large as all nearby values, the secant line from $(c, f(c))$ to a nearby point $(c+h, f(c+h))$ with $h>0$ must have slope $\le 0$ — you are looking rightward and downward or flat. The secant to a point on the left, $h<0$, must have slope $\ge 0$ — reading the quotient $\frac{f(c+h)-f(c)}{h}$ with a negative denominator flips the sign of a numerator that is still $\le 0$. So as $h \to 0^+$ the difference quotients are squeezed toward being $\le 0$, and as $h \to 0^-$ they are squeezed toward being $\ge 0$. If $f$ is differentiable at $c$, both one-sided limits must equal the same number $f'(c)$. A number that is simultaneously $\le 0$ (as a limit from the right) and $\ge 0$ (as a limit from the left) can only be $0$. This is the entire proof, compressed into one sentence; the formal version below just makes each sign statement rigorous using the definition of limit. The picture also reveals immediately why "interior" matters: at a right endpoint of the domain, there is no room to test $h>0$, so only one of the two sign constraints exists, and there is nothing to force the slope to zero.

Proof technique used: direct construction of one-sided sign inequalities from the defining local inequality, combined with the order-preserving (squeeze-type) property of limits — no appeal to any theorem built on top of Fermat's theorem itself (such as the Mean Value Theorem), so the argument is non-circular.

Example 1: A cubic with an asymmetric interior maximum. Let $f(x) = -x^3 + 3x^2 + 9x - 2$ on the interval $[-3, 5]$. We locate all candidates for global extrema. Since $f$ is a polynomial, it is differentiable everywhere, in particular on the open interval $(-3,5)$, so Fermat's theorem applies to any interior local extremum. Compute

Setting $f'(x) = 0$ gives $x = 3$ and $x = -1$, both interior to $[-3,5]$, so both are candidates permitted by the corollary above. Evaluate:

Interesting asymmetry: the global maximum value $25$ is attained both at the interior critical point $x=3$ and at the endpoint $x=-3$ — a tie between an interior critical point and a boundary point, which is exactly the scenario the corollary anticipates by requiring us to check both types of candidates. The global minimum value $-7$ is likewise attained both at interior critical point $x=-1$ and at endpoint $x=5$. Fermat's theorem correctly predicted that if an extremum were strictly interior, it would show up among $\{-1,3\}$ — and indeed both interior extrema (which are local, at least) occur exactly there. Direct sign analysis of $f'$ confirms $x=-1$ is a local minimum ($f'$ changes $- \to +$... here $f'(-2) = -3(4)+6(-2)+9=-12-12+9=-15<0$ and $f'(0)=9>0$, confirming a local min at $-1$) and $x=3$ is a local maximum ($f'(2) = -3(4)+12+9 = 9 > 0$, $f'(4) = -48+24+9=-15<0$, confirming a local max at $3$).

Example 2: A function differentiable at the extremum but not elsewhere — testing the hypotheses precisely. Let

restricted to the interior point $c=0$ of $\mathbb{R}$. This is a genuinely delicate case because $f$ oscillates infinitely often near $0$ yet is differentiable there. First confirm differentiability at $0$ directly from the definition:

since $|h \sin(1/h)| \le |h| \to 0$. So $f'(0) = 1/2 \ne 0$. Does this contradict Fermat's theorem? No — because $f$ does not have a local extremum at $0$. Indeed for small $h>0$, $f(h) \approx h/2 > 0 = f(0)$, and for small $h<0$, $f(h) \approx h/2 < 0 = f(0)$; more carefully, since $|h^2\sin(1/h)| \le h^2 = o(h)$, the term $h/2$ dominates the sign of $f(h)-f(0)$ for small $h\ne0$, so $f(h) > f(0)$ for small $h>0$ and $f(h) < f(0)$ for small $h<0$. Thus $f$ is strictly increasing through $0$ in a suitable sense and has no local extremum at $c=0$ at all — consistent with $f'(0) \ne 0$. This example is instructive precisely because it shows Fermat's theorem is a one-directional implication: differentiability plus interior-ness alone say nothing; the extremum hypothesis is what does the work, and here that hypothesis genuinely fails, so there is no contradiction.