The Mean Value Theorem

Q: What is the Mean Value Theorem?

It says that if a function is continuous on a closed interval and differentiable inside it, then at some point inside, the slope of the tangent equals the slope of the secant through the two ends. The instantaneous rate equals the average rate somewhere.

Q: What is the difference between Rolle’s Theorem and the Mean Value Theorem?

Rolle’s Theorem is the special case where the function has the same value at both ends, so the secant is horizontal and the tangent at the promised point is horizontal too. The Mean Value Theorem tilts the picture to any secant.

Q: Why does the Mean Value Theorem need continuity and differentiability?

Continuity on the closed interval lets the extreme values exist, and differentiability inside lets a tangent exist at them. Without them there may be no tangent with the right slope, as for the absolute value function on an interval that contains its corner.

Q: How do you find the value of c?

Compute the secant slope, the change in the function divided by the change in x, then solve the equation that sets the derivative equal to it. Keep the solutions that lie strictly between the two ends; there may be more than one.

Q: Is the point c in the Mean Value Theorem unique?

No. The theorem guarantees at least one. The function x cubed minus x squared on the interval from minus 1 to 2 has two such points, and sine on the interval from 0 to 3 pi has three horizontal tangents.

Q: What does the Mean Value Theorem say about speed?

A moving body must have an instantaneous speed equal to its average speed at some moment. A car that goes 176 meters in 8 seconds read exactly 22 meters per second at some instant, and a truck that averaged 115 kilometers per hour was at 115.

Q: Why does a zero derivative mean a constant function?

For any two points of the interval, the Mean Value Theorem says the change in the function equals the derivative somewhere between them times the distance. If the derivative is zero everywhere, the change is zero, so all values are equal.

Q: How is the Mean Value Theorem used to prove inequalities?

If the derivative stays between two numbers, the change in the function is between those numbers times the length of the interval. Since the derivative of sine is at most 1 in size, the sine of b minus the sine of a is at most b minus a.

Updated 21 October 2026 17 min read Intermediate Baca dalam Bahasa Indonesia

Calculus Proof Computing

Quick answer

If f is continuous on [a,b] and differentiable on (a,b), the Mean Value Theorem says that some point c in (a,b) has f'(c)=(f(b)-f(a))/(b-a): the tangent there is parallel to the secant through the two ends. Somewhere the instantaneous rate of change equals the average rate over the whole interval. Rolle's Theorem is the case f(a)=f(b), where f'(c)=0.

Key takeaways

Rolle's Theorem: if 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); the graph has a horizontal tangent between the two equal values.
Mean Value Theorem: under the same two hypotheses, f'(c)=(f(b)-f(a))/(b-a) for some c in (a,b); it is proved by applying Rolle's Theorem to the vertical gap between the graph and the secant.
Both hypotheses matter, and differentiability is needed only inside the interval, not at the ends: √(1-x^2) on [-1,1] qualifies, while |x| on [-1,1] has a corner and no horizontal tangent.
Average speed is an instantaneous speed somewhere: a car that goes 176 m in 8 s must read 22 m/s at some instant, and a truck that covers 230 km in 2 hours on a 100 km/h road was speeding.
Corollaries: if f'=0 on an interval then f is constant there, and if f'=g' then f=g+C; so every antiderivative of a function is one of them plus a constant, and one point fixes C.
Uses: an equation has at most one more real solution than its derivative has zeros, and a bound m\le f'\le M gives m(b-a)\le f(b)-f(a)\le M(b-a).

What is Rolle’s Theorem?

Rolle's Theorem says that if y=f(x) is continuous on the closed interval [a,b], differentiable at every point of (a,b), and f(a)=f(b), then there is at least one number c in (a,b) at which f'(c)=0. As the graph suggests, a differentiable curve that crosses a horizontal line at two points has a horizontal tangent somewhere between them. There may be just one such point, or several.

Proof. Because f is continuous, the Extreme Value Theorem gives an absolute maximum and an absolute minimum on [a,b], and each is taken at an interior point where f' is zero, at an interior point where f' does not exist, or at an endpoint. The hypothesis rules out the second possibility. If the maximum or the minimum is at an interior point c, then f'(c)=0 by the First Derivative Theorem, and c is the point we want. If both are at the endpoints, then since f(a)=f(b) the largest and the smallest value are equal, so f is constant, f'(x)=0 everywhere inside, and c can be any point of (a,b). ∎

The hypotheses are essential. If they fail at even one point, the graph may have no horizontal tangent at all: - discontinuous at an endpoint: f(x)=x for 0\le x<1 and f(1)=0 has f(0)=f(1)=0, but its slope is 1 everywhere inside; - discontinuous at an interior point: a graph that jumps over the gap leaves both pieces sloping the same way; - continuous but not differentiable at an interior point: |x| on [-1,1] has f(-1)=f(1) and a corner at 0, with slopes -1 and then 1, never 0; x^2/3 on [-1,1] has a cusp at 0.

The opposite also needs care: a failed hypothesis does not make the conclusion false, only unpromised. Choose a case below and see which hypothesis it breaks.

Interactive: Rolle’s Theorem, case by case

Seven functions with f(a) = f(b): three where the theorem gives horizontal tangents, and four where a jump, a corner or a cusp breaks a hypothesis and no tangent is horizontal.

How does Rolle’s Theorem count the solutions of an equation?

Rolle's Theorem turns a statement about zeros of f into a statement about zeros of f': between any two zeros of f there is a zero of f'. So if f' has k zeros, f cannot have more than k+1.

Example 1. Show that x^3+3x+1=0 has exactly one real solution. Let f(x)=x^3+3x+1, which is continuous. Since f(-1)=-3 and f(0)=1, the Intermediate Value Theorem says the graph crosses the x-axis somewhere in (-1,0), so there is at least one solution. Now suppose there were two zeros x=a and x=b. Rolle's Theorem would give a point c between them with f'(c)=0. But f'(x)=3x^2+3 is never zero, because it is always at least 3. So there cannot be two zeros, and there is exactly one.

The same reasoning counts roots in general. A cubic has a quadratic derivative with at most two zeros, so a cubic has at most three real zeros. For a polynomial with n different real zeros, every one of the n-1 gaps between neighbours holds a zero of f', and f' has degree n-1, so those are all of its zeros: exactly one in each gap. Choose the zeros below and see it happen; the point with the horizontal tangent is always caught between two zeros.

Interactive: between two zeros, a zero of the derivative

Choose the zeros of a polynomial and see its graph, its derivative and the point with a horizontal tangent that Rolle’s Theorem puts between every two consecutive zeros.

What does the Mean Value Theorem say?

The Mean Value Theorem is a slanted version of Rolle's Theorem: if y=f(x) is continuous on [a,b] and differentiable on (a,b), then there is at least one point c in (a,b) at which (f(b)-f(a))/(b-a)=f'(c). Geometrically, the left side is the slope of the secant line through A(a,f(a)) and B(b,f(b)), and the right side is the slope of the tangent at c: somewhere between a and b the curve has a tangent line parallel to the secant. When f(a)=f(b) the secant is horizontal and this is Rolle's Theorem.

Proof. The secant line is the graph of g(x)=f(a)+(f(b)-f(a))/(b-a)(x-a). The vertical distance between the graph of f and the secant is h(x)=f(x)-g(x) h(x)=f(x)-f(a)-(f(b)-f(a))/(b-a)(x-a). The function h is continuous on [a,b] and differentiable on (a,b) because f and g are, and h(a)=h(b)=0 because both graphs pass through A and B. So Rolle's Theorem gives a point c in (a,b) with h'(c)=0. Differentiating h and putting x=c: h'(c)=f'(c)-(f(b)-f(a))/(b-a)=0, which is the equation of the theorem. ∎

What it needs, and what it does not. Continuity on the closed interval and differentiability inside are all that the proof used. In particular f need not be differentiable at a or at b: one-sided continuity is enough. The function f(x)=√(1-x^2) on [-1,1] has vertical tangents at both ends, yet it satisfies the theorem, with c=0.

What it does not say. The theorem does not say there is only one point, and it gives no formula for c; it says only that at least one exists. Type a function and an interval below, or pick an example. The orange line is the secant, the green lines are the tangents at the points the theorem promises, and the two cases where a hypothesis fails show what the theorem does not cover.

Interactive: the Mean Value Theorem

Type a function and the ends of an interval, or pick a textbook example, and see the secant through the two ends, every tangent line parallel to it, and whether the two hypotheses (continuous, differentiable inside) hold.

How do you find the point c?

To find c, compute the slope of the secant m=(f(b)-f(a))/(b-a), solve f'(c)=m, and keep the solutions that lie inside the open interval (a,b). The theorem guarantees that the list is not empty when the hypotheses hold.

Example 2. f(x)=x^2 on [0,2]. The slope is (4-0)/(2-0)=2 and f'(c)=2c=2, so c=1.

Example 3. f(x)=x^2+2x-1 on [0,1]. The slope is (2-(-1))/(1-0)=3, and f'(c)=2c+2=3 gives c=\frac12.

Example 4. f(x)=x^2/3 on [0,1]. The slope is 1 and f'(c)=\frac23c^-1/3=1 gives c^1/3=\frac23, so c=(8)/(27). The cusp is at 0, an endpoint, so f is differentiable inside and the theorem applies.

Example 5. f(x)=x^3-x^2 on [-1,2]. The slope is (4-(-2))/(3)=2, and f'(c)=3c^2-2c=2 gives 3c^2-2c-2=0 c=(1±√7)/(3). Both c≈-0.549 and c≈1.215 lie in (-1,2), so there are two points. The theorem said "at least one".

The two means. For f(x)=\frac1x on an interval [a,b] of positive numbers, the slope is -\frac1ab and f'(c)=-\frac1c^2, so c=√(ab), the geometric mean of a and b. For f(x)=x^2 it is 2c=a+b, so c=(a+b)/(2), the arithmetic mean. Indeed for every quadratic px^2+qx+r the slope of the secant is p(a+b)+q and f'(c)=2pc+q, so c is always the midpoint, and it is the only such point.

When c cannot be written down. For f(x)=xe^x on [0,1] the slope is e and the equation is (1+c)e^c=e. It has exactly one solution in (0,1), but no elementary formula; a computer finds c≈0.5571. The theorem is valuable because it promises c without needing to compute it.

What does the theorem say about speed and average rates?

Read as a statement about rates, the Mean Value Theorem says that the average rate of change over an interval is the instantaneous rate at some interior point. The number (f(b)-f(a))/(b-a) is the average change of f over [a,b], and f'(c) is an instantaneous change; the theorem says they meet. It applies to anything whose rate is a derivative, as in the motion along a line of the rates article.

Example 6. A car accelerating from rest takes 8 s to go 176 m. Its average velocity is (176)/(8)=22 m/s, so at some moment during the acceleration the speedometer must read exactly 22 m/s, which is 22×3.6=79.2 km/h. If the car has s=2.75t^2, which does match 176 m at t=8, then v=s'=5.5t equals 22 at t=4 s.

A speeding ticket. A truck driver hands in a tollbooth ticket showing that she covered 230 km of a toll road with a 100 km/h limit in 2 hours. The average speed is 115 km/h, so by the theorem the truck was doing exactly 115 km/h at some instant, above the limit: a citation is justified without anyone having seen her speed.

More of the same. - A thermometer taken from a freezer at -19 °C into boiling water at 100 °C and reading it after 14 s rose (119)/(14)=8.5 °C per second on average, so at some moment the mercury was rising at exactly 8.5 °C/s. - A trireme that covered 184 sea miles in 24 hours averaged 7.67 knots, so at some point its speed exceeded 7.5 knots. - A marathoner who ran 42 km in 2.2 hours averaged 19.09 km/h. Her speed was 19.09 km/h at some instant, by the theorem. It starts at 0 and ends at 0, and a speed that changes continuously cannot skip values (the Intermediate Value Theorem), so it passed 18 km/h once on the way up and again on the way down: at least twice.

What it does not say. The theorem promises an instant, not that the speed stayed at the average, and it does not say when. It is an existence statement, and that is exactly what a proof or a legal argument needs.

What follows when two functions have the same derivative?

The first question the theorem answers is which functions have a zero derivative: only constant ones. Corollary 1: if f'(x)=0 at every point of an open interval (a,b), then f(x)=C for all x in (a,b), where C is a constant.

Proof. Take any two points x_1<x_2 in (a,b). The function f is differentiable on [x_1,x_2], hence continuous, so the theorem gives a point c between them with (f(x_2)-f(x_1))/(x_2-x_1)=f'(c)=0. Then f(x_2)-f(x_1)=0 and f(x_1)=f(x_2). Since x_1 and x_2 were arbitrary, f has one value on the whole interval. ∎

The second question is how two functions with the same derivative are related. Corollary 2: if f'(x)=g'(x) at every point of an open interval (a,b), then there is a constant C with f(x)=g(x)+C for all x in (a,b). Apply Corollary 1 to h=f-g, whose derivative is h'=f'-g'=0 on (a,b), so h=C there. Both corollaries remain true for infinite intervals. Geometrically, the graphs of functions with the same derivative differ only by a vertical shift, as the parabolas y=x^2+C do for the derivative 2x.

This is what makes antiderivatives well behaved. Once one function g has the derivative f' you want, every such function is g(x)+C, and a single point of the graph fixes C.

Example 7. Find the function f(x) whose derivative is \sin x and whose graph passes through (0,2). The function g(x)=-\cos x has g'(x)=\sin x, so f(x)=-\cos x+C. From f(0)=-\cos0+C=2 we get C=3, and f(x)=-\cos x+3.

Example 8, free fall. A body falls from rest with acceleration 9.8 m/s², and the position s is measured downward from the rest position. The velocity v has derivative 9.8, and so does 9.8t, hence v(t)=9.8t+C, and v(0)=0 gives C=0: v(t)=9.8t. The position s has derivative 9.8t, and so does 4.9t^2, hence s(t)=4.9t^2+C, and s(0)=0 gives s(t)=4.9t^2, until the body hits the ground. This is the same fall as in the free fall and throws of the rates article, worked out backward from the acceleration.

Pick a derivative below and move the point the graph must pass through. The other curves are the same graph shifted, and only one of them goes through the point.

Interactive: functions with the same derivative

Pick a derivative and see that every function with that derivative is one curve shifted up or down by a constant C; move the point the graph must pass through and watch C change.

How does the theorem prove inequalities and bounds?

If m\le f'(x)\le M for every x in (a,b), and f is continuous on [a,b], then m(b-a)\le f(b)-f(a)\le M(b-a). The reason is one line: the theorem gives (f(b)-f(a))/(b-a)=f'(c) for some c, and f'(c) is between m and M. Knowing how fast a function can change tells how far it can move.

Example 9. If |f'(x)|\le1 for 1\le x\le4, then -3\le f(4)-f(1)\le3, because b-a=3.

Example 10, the sine inequality. For any a and b, |\sin b-\sin a|\le|b-a|. The derivative of \sin is \cos, and |\cos|\le1, so |\sin b-\sin a|=|\cos c| |b-a|\le|b-a|. In the same way |\cos x-1|\le|x| for all x: apply the theorem to \cos t on the interval from 0 to x, where |\sin c|\le1.

Example 11, an estimate. For f(x)=√(1+x) on [0,0.1] the derivative f'(x)=\frac12√(1+x) decreases from 0.5 to \frac12√(1.1)≈0.4767, so 0.4767·0.1\le f(0.1)-f(0)\le0.5·0.1, and 1.0477\le√(1.1)\le1.05. The true value is 1.0488. This is the same estimate as the linearization 1+(0.1)/(2)=1.05, now with a guaranteed range on both sides, which linearization alone does not give.

Example 12. If f'(x)\le3 for all x and f(0)=1, the largest f(2) can be is 1+3·2=7, since f(2)-f(0)\le3·2. And \frac12\le\frac11+x^2\le1 on (0,1) means \frac12\le\arctan1\le1 (it is \fracπ4≈0.785).

How do you apply the theorem in code?

In code, finding c means solving f'(c)=m on the open interval, either exactly or by scanning for sign changes of f'(x)-m and refining them by bisection. SymPy does it exactly: sp.solveset(sp.diff(f, x) - slope, x, sp.Interval.open(a, b)) returns every solution inside, and an empty set is the honest answer when a hypothesis fails. The scan has one trap that the theorem warns about: at a corner the derivative jumps across the level m, and a sign change there is not a zero, so each candidate has to be checked by evaluating f'-m at the refined point.

Python and SymPy: the points c, exactly

import sympy as sp

x = sp.symbols('x', real=True)


def mvt_points(f, a, b):
    """The slope of the secant, and every c in (a, b) where f'(c) equals it."""
    slope = sp.simplify((f.subs(x, b) - f.subs(x, a)) / (b - a))
    return slope, sp.solveset(sp.diff(f, x) - slope, x, sp.Interval.open(a, b))


examples = (
    (x**2, 0, 2),                      # Example 2
    (x**3 - x**2, -1, 2),              # Example 5: two points
    (1 / x, 1, 4),                     # the geometric mean sqrt(1 * 4)
    (sp.sqrt(1 - x**2), -1, 1),        # not differentiable at the ends, still c = 0
    (sp.sin(x), 0, 3 * sp.pi),         # Rolle: three horizontal tangents
    (sp.Abs(x), -1, 2),                # a corner: a hypothesis fails and there is no c
)
for f, a, b in examples:
    print(f, [a, b], '->', mvt_points(f, a, b))
# x**2 [0, 2] -> (2, {1})
# x**3 - x**2 [-1, 2] -> (2, {1/3 - sqrt(7)/3, 1/3 + sqrt(7)/3})
# 1/x [1, 4] -> (-1/4, {2})
# sqrt(1 - x**2) [-1, 1] -> (0, {0})
# sin(x) [0, 3*pi] -> (0, {pi/2, 3*pi/2, 5*pi/2})
# Abs(x) [-1, 2] -> (1/3, EmptySet)

JavaScript: scan and bisect, without a CAS

const mvt = (f, df, a, b, n = 4000) => {
  const slope = (f(b) - f(a)) / (b - a)
  const g = (x) => df(x) - slope                              // the points c are the zeros of g
  const cs = []
  for (let i = 0; i < n; i++) {
    let lo = a + ((b - a) * i) / n
    let hi = a + ((b - a) * (i + 1)) / n
    if (g(lo) === 0 && i > 0) cs.push(lo)                     // a zero that falls exactly on a grid point
    else if (g(lo) * g(hi) < 0) {                             // g changes sign: bisect
      for (let k = 0; k < 60; k++) {
        const mid = (lo + hi) / 2
        if (g(lo) * g(mid) <= 0) hi = mid
        else lo = mid
      }
      const c = (lo + hi) / 2
      if (Math.abs(g(c)) < 1e-6) cs.push(c)                   // a corner makes g jump across 0: that is not a zero
    }
  }
  return { slope, cs }
}
mvt((x) => x * x, (x) => 2 * x, 0, 2)                                        // slope 2, cs [1]
mvt((x) => x ** 3 - x ** 2, (x) => 3 * x * x - 2 * x, -1, 2)                 // slope 2, cs [-0.5486, 1.2153]
mvt((x) => x * Math.exp(x), (x) => (1 + x) * Math.exp(x), 0, 1)              // slope e = 2.7183, cs [0.5571]
mvt((x) => 1 / x, (x) => -1 / (x * x), 1, 4)                                 // slope -0.25, cs [2]
mvt(Math.abs, Math.sign, -1, 2)                                              // slope 0.3333, cs []: the corner at 0 is rejected

C++

#include <iostream>
#include <cstdio>
#include <cmath>
using namespace std;

double f(int k, double x) {
  if (k == 0) return x * x * x - x * x;       // Example 5 on [-1, 2]
  if (k == 1) return x * exp(x);              // no formula for c, on [0, 1]
  return 1 / x;                               // the geometric mean, on [1, 4]
}
double df(int k, double x) {
  if (k == 0) return 3 * x * x - 2 * x;
  if (k == 1) return (1 + x) * exp(x);
  return -1 / (x * x);
}

void mvt(int k, double a, double b) {
  double slope = (f(k, b) - f(k, a)) / (b - a);
  int n = 2000;
  for (int i = 0; i < n; i++) {
    double lo = a + (b - a) * i / n;
    double hi = a + (b - a) * (i + 1) / n;
    if ((df(k, lo) - slope) * (df(k, hi) - slope) < 0) {     // f' - slope changes sign: bisect
      for (int j = 0; j < 60; j++) {
        double mid = (lo + hi) / 2;
        if ((df(k, lo) - slope) * (df(k, mid) - slope) <= 0) hi = mid; else lo = mid;
      }
      double c = (lo + hi) / 2;
      printf("slope=%.4f c=%.4f f'(c)=%.4f\n", slope, c, df(k, c));
    }
  }
}

int main() {
  mvt(0, -1, 2);
  mvt(1, 0, 1);
  mvt(2, 1, 4);
  printf("%.4f %.4f\n", sqrt(1.0 * 4), (1.0 + 4) / 2);   // the geometric and the arithmetic mean of 1 and 4
  return 0;
}

/* output:
slope=2.0000 c=-0.5486 f'(c)=2.0000
slope=2.0000 c=1.2153 f'(c)=2.0000
slope=2.7183 c=0.5571 f'(c)=2.7183
slope=-0.2500 c=2.0000 f'(c)=-0.2500
2.0000 2.5000
*/

Where does the Mean Value Theorem come from?

The theorem is older than the definition of the derivative it is stated with, and it was proved first for polynomials.

Parameshvara, 15th century. The Indian astronomer Parameshvara of the Kerala school is credited with a special case of the theorem, for the sine function.
Rolle, 1691. Michel Rolle showed, for polynomial equations, that between two roots there lies a root of the derivative, as a step in a method for separating the roots. He was a critic of the new infinitesimal calculus, and the name Rolle's theorem was given to the result in the 19th century (by Giusto Bellavitis, in 1846).
Lagrange, 1797. Joseph-Louis Lagrange stated the Mean Value Theorem and used it as the base of his attempt to build calculus on power series. Its other name, the Lagrange mean value theorem, is from this.
Cauchy, 1823. Augustin-Louis Cauchy, with the derivative defined as a limit, gave the proof by Rolle's Theorem used in books today, and a generalization to two functions, (f(b)-f(a))/(g(b)-g(a))=(f'(c))/(g'(c)), from which the rule of l'Hôpital for limits follows.

What mistakes do people make with the theorem?

The most common mistakes with the Mean Value Theorem are the nine below, each with the correct statement.

Mistake	Correct
❌ There is exactly one point c	The theorem says at least one: x^3-x^2 on [-1,2] has two, and \sin x on [0,3π] has three horizontal tangents.
❌ Use it on \|x\| over [-1,2]	There is a corner at 0, so f is not differentiable inside; the slope of the secant is \frac13 and no tangent has that slope.
❌ Use it on \frac1x over [-1,1]	f is not continuous there, not even defined at 0, so nothing is promised.
❌ f must be differentiable at the endpoints	Only continuity at the ends is needed: √(1-x^2) on [-1,1] works, with c=0.
❌ The theorem says f(c) is the average of the values of f	It says f'(c) is the average rate, (f(b)-f(a))/(b-a); averages of values belong to integrals.
❌ f'=0 on (-1,0)∪(0,1), so f is constant	The set must be one interval: f(x)=(x)/(\|x\|) has f'=0 wherever defined and is -1 and then 1.
❌ c is always the midpoint (a+b)/(2)	Only for quadratics: x^3 on [0,2] has c=(2)/(√3)≈1.155, not 1.
❌ If f'=g' then f=g	They differ by a constant: x^2 and x^2+3 have the same derivative.
❌ An average speed of 115 km/h means the speed was always 115	The theorem gives at least one instant at that speed, not all of them.

Practice: test your understanding

These questions mix everything above. A wrong answer costs nothing here: read the hint and try again.

True or false?

Decide whether each statement is True or False.

Rolle's Theorem is the Mean Value Theorem for a function with f(a)=f(b).
The Mean Value Theorem needs f to be differentiable at a and at b.
If f is not continuous on [a,b], the conclusion can fail.
There is exactly one point c in the conclusion.
If f'=0 on an interval, then f is constant on it.
If f'=0 on (-1,0)∪(0,1), then f is constant on that set.

Select all that apply

Choose all the functions that satisfy both hypotheses of the Mean Value Theorem on [-1,1].

x^2
|x|
√(1-x^2)
\frac1x
x^1/3
x^3-x

Find c

Find the value of c in the Mean Value Theorem for f(x)=2x^2-x on [1,4].

An average speed

A train travels 270 km in 3 hours. At some instant its speed is exactly how many km/h?

What the theorem promises for |x|

What does the Mean Value Theorem say about f(x)=|x| on [-1,2]?

nothing: f is not differentiable at 0, inside the interval
there is a point c with f'(c)=\frac13
there is a point c with f'(c)=0
nothing: f is not continuous

Rolle’s Theorem

Find the c of Rolle's Theorem for f(x)=x^2-6x+5 on [1,5].

The same derivative

f'(x)=6x-2 for all x and f(0)=5. Find f(1).

The toll road

A driver covers 230 km of a road with a limit of 100 km/h in 2 hours. Why was she speeding?

her average speed was 115 km/h, so at some instant she went exactly 115 km/h
she was above 100 km/h at every instant
her speed was 100 km/h at the end
the average of two speeds is always above the limit

A sine antiderivative

g'(x)=\cos x and g(0)=1. Find g\left(\fracπ2\right).

How many solutions?

How many real solutions does x^5+2x+3=0 have?

exactly one
none
exactly two
exactly five

A bound

f'(x)\le3 for all x and f(0)=1. What is the largest value f(2) can have?

Zeros of the derivative

A polynomial has the four real zeros -2,1,4,7. How many real zeros does its derivative have?

Summary: the Mean Value Theorem at a glance

Rolle's Theorem: f continuous on [a,b], differentiable on (a,b), f(a)=f(b) give some c with f'(c)=0.
Mean Value Theorem: the same two hypotheses give some c with f'(c)=(f(b)-f(a))/(b-a), a tangent parallel to the secant.
Proof: apply Rolle's Theorem to h(x)=f(x)-f(a)-(f(b)-f(a))/(b-a)(x-a).
Watch for: corners, cusps and poles inside the interval; the point c is not unique and has no formula in general.
Rates: the average rate over an interval is the instantaneous rate at some interior instant.
Corollaries: f'=0 on an interval means constant; f'=g' means f=g+C.
Bounds: m\le f'\le M gives m(b-a)\le f(b)-f(a)\le M(b-a).
Next: the sign of f' on an interval tells where f rises and falls, the subject of the next article.
Code: sp.solveset(sp.diff(f, x) - slope, x, sp.Interval.open(a, b)), or a scan with bisection that rejects jumps.

Glossary: key terms

Rolle’s Theorem: The theorem that a function continuous on [a, b], differentiable on (a, b) and equal at the two ends has a point inside where the derivative is zero.
Mean Value Theorem: The theorem that a function continuous on [a, b] and differentiable on (a, b) has a point inside where the tangent is parallel to the secant through the ends.
Secant line: The line through two points of a graph; its slope is the average rate of change between them.
Average rate of change: The change in the value of a function divided by the change in the variable over an interval: the slope of the secant.
Antiderivative: A function whose derivative is a given function; on an interval, any two antiderivatives differ by a constant.
Corollary: A statement that follows directly from a theorem; the zero-derivative and equal-derivative results follow from the Mean Value Theorem.

Frequently asked questions

What is the Mean Value Theorem?