The Mean Value Theorem
Rolle’s Theorem: Let f be a function such that:
1. It is continuous on the closed interval [a, b].
2. It is differentiable on the open interval (a, b).
3. f (a) = f (b).
Then there is (at least) a number c in (a, b) such that f ′(c) = 0.
ex. Show that the function f (x) = x4 + 32x + k has at most 2 real roots
(where k is any constant).
Solution: Suppose there are 3 (or more) real roots, denote them x1, x2,
and x3. Without the loss of generality, assume that x1 < x2 < x3. Since
f (x1) = f (x2) = 0, by the Rolle’s Theorem there is a number c1 in (x1, x2)
such that f ′( c1) = 0. And since f (x2) = f (x3) = 0, there is a second
number c2 in (x2, x3), which is necessarily different than c1, such that
f ′( c2) = 0. But f ′(x) = 4x3 + 32 = 4(x + 2)(x2 − 2x + 4) equals zero at
only one real number, when x = −2. (It also has two complex zeroes.)
Therefore, c2 cannot possibly exist – a violation of the Rolle’s
Theorem unless there are at most 2 roots, x1 and x2.
Note: The above only says that f (x) can have at most 2 real roots. It doesn’t
say that f (x) will actually have 2 real roots. In fact, it can be easily verified
that for any constant k > 48, f (x) will not have any real root at all.
The “slanted” version of the Rolle’s Theorem is the Mean Value Theorem.
The Mean Value Theorem: Let f be a function such that:
1. It is continuous on the closed interval [a, b].
2. It is differentiable on the open interval (a, b).
Then there is (at least) a number c in (a, b) such that
f ′(c) =
f (b) − f ( a )
b−a
Comment: Notice that the left side is the instantaneous rate of change / slope
of the tangent line at c, while the right side is the average rate of change /
slope of the secant line from a to b. Thus the theorem guarantees that there
is at least one place in (a, b) where the tangent line there is parallel to the
secant line connecting the points (a, f (a)) and (b, f (b)) on the curve y = f (x).
ex. Find the value(s) of c such that the Mean Value Theorem is satisfied for
the function f (x) = x2/3 on the interval [0, 1].
First, let’s note that f is indeed continuous on [0, 1], and is
differentiable on (0, 1), so the prerequisites for the M.V.T. are met.
(The function is not differentiable at x = 0, but it nevertheless meets
the condition of being differentiable on the open interval (0, 1).)
The average rate of change from x = 0 to x = 1 is
f (1) − f (0) 1
= =1
1− 0
1
The Mean Value Theorem says that there is at least one number c on
(0, 1) such that f ′(c) = 1. So set f ′(x) = 1 and solve:
f ′( x) =
1
3
3
=
x 2
2 −1 / 3
x
=1
3
3
2
x=
3
→
x −1 / 3 =
3
2
3
8
2
x=  =
27
3
Therefore, c = 8/27, which is indeed within the interval (0, 1).
ex. Suppose it took 14 seconds for a thermometer to rise from −19°C to
100°C. Show that sometime between t = 0 and t = 14 (sec) the mercury is
rising at the exact rate of 8.5°C/sec.
The average rate of change of the temperature function f , between t =
0 and t = 14 is
f (14) − f (0) 100 − ( −19) 119
=
= 8 .5
=
14 − 0
14
14
Since the temperature, as a function, is continuous and differentiable
everywhere, the Mean Value Theorem applies. Therefore, by the
M.V.T., there is, at least, one moment within the given time interval
when the mercury is rising at the exact rate of 8.5°C/sec. (Notice that
we don’t even need to know what the function was to prove the above.)
x , 0 ≤ x < 1
ex. Consider the function f ( x) = 
x =1
0 ,
The function is differentiable on (0, 1), and f (0) = f (1) = 0, but
nowhere within (0, 1) is f ′(x) = 0. In fact, f ′(x) = 1 everywhere on (0,
1). Why isn’t this fact a violation of the Rolle’s Theorem (and
therefore violating the M.V.T. as well)?
Answer: Because f is not continuous on [0, 1]. It has a jump
discontinuity at x = 1. As a result it does not satisfy one of the
prerequisites of the Rolle’s Theorem. Therefore, the theorem does not
apply.
ex. Consider the function f (x) = x2/3 in the first example. There is no value
of c that satisfies the M.V.T. on the interval [−1, 1]. This is not a violation
of the M.V.T., either. Because f is not differentiable at x = 0, so it is not
differentiable on the entire open interval (−1, 1), therefore the M.V.T. does
not apply.
ex. Suppose that f is a differentiable function such that f (4) = 5. If, for all
values of x, −3 ≤ f ′(x) ≤ 2, then what range of values can f (10) have?
Since −3 ≤ f ′(x) ≤ 2 for all x, by the Mean Value Theorem the average
rate of change of f on any interval has to be bounded between −3 and
2 as well. Therefore, on the interval [4, 10], which has a length of 6,
the values of f can change between −18 (if the average rate of change
is −3, the minimum) and 12 (if the rate is 2, the maximum). Hence
the range of values f (10) can possibly obtain, starting at f (4) = 5, is
between f (4) – 18 = −13 and f (4) + 12 = 17, i.e. −13 ≤ f (10) ≤ 17.
Some consequences of the Mean Value Theorem
Theorem: (Functions with zero derivatives are constant.) If f ′(x) = 0 for
all x in an interval (a, b), then f is constant on (a, b).
We already know that all constant functions have zero derivatives.
This theorem states that they are all the functions with such property.
There are no non-constant functions that have zero derivatives.
Corollary: (Functions with the same derivative differ from one another
only by a constant.) If f ′(x) = g′(x) for all x in an interval (a, b), then f − g
is a constant on (a, b), i.e., f (x) = g(x) + c, where c is a constant.
Note that (infinitely) many functions can share the same derivative.
For example, f1(x) = x2, f2(x) = x2 + 1, f3(x) = x2 + 2, and f4(x) = x2 − π
all have the same derivative, 2x. The above result says that all
functions with the same derivative differ by a constant with one
another. Therefore, knowing one such function gives us all of them.
For the above example, functions with derivative 2x have the general
form f (x) = x2 + C, where C is any number.
This is an important fact that will become very useful later. It enables us to
find all antiderivatives of a function by just finding one antiderivative.