5.7 - Rolle’s Theorem, Mean Value Theorem
All right, we are done with the optimization problems. Now we move on to a couple of very
famous (and useful) theorems in calculus.
Say f(x) is a continuous function on a closed interval [a;b] and di▯erentiable on (a;b). Then if
f(a) = f(b), there must be some point c, a ▯ c ▯ b such that f (c) = 0.
So what does that mean? It says that if you have a nice function where the end points give you
the same value, then somewhere in between, there must be a horizontal tangent. Think of
f(x) = x , for example. Well, f(▯2) = f(2) = 4. Somewhere between [▯2;2], there is a number c
such that f (c) = 0. In this case, c = 0.
Why is it that we can guarantee this c exists? Well, let’s just think about the case that
f(a) = f(b) = 0 (if the function isn’t like this, just translate it so that it does!). Notice that if
f(x) = 0 (the zero function), this is a bit trivial (f (x) = 0, and we have every point between a
and b gives the derivative to be zero!). So, at some point, f(x) has to have a value other than zero.
Say, for argument’s sake, that f(x) goes up. So, this means that f(x) > 0 at some point in (a;b).
Now, we have a continuous function on [a;b], a closed interval. Remind you of anything? Extreme
Value Theorem perhaps? Remember, Extreme Value Theorem says if you have a continuous
function on a closed interval, there must be an absolute max and an absolute min. Well, i don’t
care about the min right now, so let’s take a look at that max. Here we have f(a) = 0 and
f(b) = 0, and at some point, we know f(x) > 0. This means that the absolute max occurs
somewhere in between a and b, namely, say, oh, at c. So, we can therefore conclude that c is in
fact a critical point, and since we’re di▯erentiable on (a;b), then this critical point is also a
stationary point, meaning f (c) = 0.
Notice this basically works the same if f(x) < 0 between a and b.
Okay, so we have this Rolle’s Theorem thing. What’s so useful about it? Well, the idea is, it
guarantees this c. Many times we could care less what the actual value of c is, just that it exists.
Sometimes, it’s actually impossible to determine c, but at least we know it’s there. However, in
your homework, you will be asked many times to verify that a c exists and then you will be asked
to go ahead and actually ▯nd that c.
Example: here’s one that we actually can ▯nd c. Let f(x) = x ▯ x ▯ 6. Find the x-intercepts
and con▯rm that f (x) = 0 for some c between the x-intercepts.
So the idea here is that you have these x-intercepts, and using the fact that f(x) = 0 at these two
places, we can apply Rolle’s Theorem.
f(x) = x ▯ x ▯ 6 = (x + 2)(x ▯ 3) = 0
so the x-intercepts are x = ▯2 and x = 3.
But ▯rst, we have to check and make sure that we can use Rolle’s Theorem. If i ask a question like
this and the answer is using Rolle’s Theorem, i expect a little checklist like this one coming up:
(1) Is f(x) continuous on [▯2;3]? Yup. It’s a polynomial! X
1 (2) Is f(x) di▯erentiable on (▯2;3)? Yup! Same reason: it’s a polynomial. X
(3) f(▯2) = f(3) = 0 X
Why do we need this checklist? Because you can’t just take Rolle’s Theorem out of thin air on
noncontinuous or nondi▯erentiable functions (and well, de▯nitely not so if f(▯2) 6= f(3), but
they’re x-intercepts and we wouldn’t be bothering with Rolle’s Theorem if they’re not). If you do
not put that checklist, or at least acknowledge the fact that f(x) is cont and di▯, points will be
taken o▯ on a tests and quizzes. Even if you are completely correct in saying Rolle’s Theorem is
how to do it, you need to verify that you can actually use Rolle’s Theorem. So, anyway, back to
We need to ▯nd the point c where f (c) = 0. Taking the derivative,
f (x) = 2x ▯ 1
and setting that to zero, we get x = 1=2. This is our c. c = 1=2 because f (1=2) = 0!
Here’s something else useful about Rolle’s Theorem. Cubic functions can have either 1 or 3 real
roots (they’ll either hit the x-axis once or 3 times, including multiplicity). Well, this is great and
all, but a big pain when you have a cubic function and can’t tell if it has any more solutions (like,
say, if you’re trying to graph it or something). Rolle’s Theorem can help you determine if there’s
another root or not. Here’s an example.
Example: Let f(x) = x + x ▯ 1. Show that f(x) has exactly one real root.
Solution: Well, for one thing, f(x) is continuous and f(0) = ▯1 and f(1) = 2, so somewhere
between x = 0 and x = 1, the function has to hit zero (Intermediate Value Theorem). So, it has at
least one root. Now we need to show that there are no more. So, let’s try a contradiction.
Suppose, for now, that there are in fact 2 real roots to this f(x), say a and b. Well, since f(x) is
continuous and di▯erentiable (it’s a polynomial!) and f(a) = f(b) = 0, then we can apply Rolle’s
Theorem, which says there is some c between a and b such that f (c) = 0. Okay... So, let’s take
the derivative of f(x) and check out this c.
f (x) = 3x + 1
This would mean that if we were to set f (x) = 0, we would have
x = ▯1=3
No way we can have that! This means no such c exists. Where’s the problem? Well, f(x) is
de▯nitely continuous and di▯erentiable, and Rolle’s Theorem can’t be wrong, so the problem must
be in the f(a) = f(b) = 0. Namely, we can’t have two roots. There’s only one.
One of the greatest uses of all of Rolle’s Theorem is that it’s used to prove a more general result:
The Mean Value Theorem.
Mean Value Theorem
if f(x) is continuous on [a;b] and di▯erentiable on (a;b), then there is some c between a and b
0 f(b) ▯ f(a)
f (c) =
b ▯ a
So here it is, explained a little. You have a function with end points a and b. Draw a line from
f(a) to f(b). What’s the slope of that line? b▯a ! So, what the MVT is saying is at some
point between a and b, the tangent line has the same slope as the line from f(a) to f(b). Notice
how Rolle’s Theorem is a speci▯c case of MVT (when f(a) = f(b), then b▯a = 0).
Consider equation of line with slope f(b)▯f(a)going through the point (a;f(a)).
f(b) ▯ f(a)
y ▯ f(a) = (x ▯ a)
b ▯ a
f(b) ▯ f(a)
y = (x ▯ a) + f(a)
b ▯ a
Now consider the function that is the di▯erence between f(x) and this line. Call this g(x).
f(b) ▯ f(a)
g(x) = f(x) ▯ ( b ▯ a (x ▯ a) + f(a))
Well, g(x) is the di▯erence of two continuous and di▯erentiable functions, which means it’s
continuous and di▯erentiable. Also, notice that
f(b) ▯ f(a)
g(a) = f(a) ▯ ( (a ▯ a) + f(a)) = f(a) ▯ (0 + f(a)) = f(a) ▯ f(a) = 0
b ▯ a
g(b) = f(b)▯( f(b) ▯ f(a)(b▯a)+f(a)) = f(b)▯(f(b)▯f(a)+f(a)) = f(b)▯f(b)+f(a)▯f(a) = 0
b ▯ a
Oh, okay! We have a cont and di▯ function where g(a) = g(b) = 0! So we can apply Rolle’s
Theorem! That means there’s some c between a and b such that
0 0 f(b) ▯ f(a)