xaktly | Calculus

Mean value theorem

Existence theorems

There is a class of important theorems in calculus which can be called existence theorems. They are gathered here for easy reference.

While a few of these may seem obvious when you read them — for sure there's no big mystery behind them — they can be helpful at solving some important problems and for proving other theorems.

They can also feel pretty esoteric the first few times you encounter them. I suggest coming back to this section once in a while to refresh your memory. You'll begin to make more connections between these important theorems and the rest of calculus and mathematics in general.

No. Theorem Description

Continuity of a function

(not really an existence theorem, but an important foundation for the others)

Intermediate value
theorem (IVT)

A continuous function cannot skip values.

Extreme value theorem

Any continuous function has a global max. & min. on a closed interval.

Fermat's theorem

f(x) = 0 at a maximum or minimum.

Rolle's theorem

Special case of the mean value theorem.

Mean value theorem (MVT)

Any continuous function takes on its average value at least once in an open interval. This simple theorem underlies much of calculus.

1. Continuity

The idea of continuity of a function is pretty easy to wrap your mind around: A function is continuous if you can draw it without lifting your pen. You just have to remember that a discontinuity like a hole is still a place where you have to lift your pen – no matter how infinitesimally small the gap. It's the mathematical statements of continuity that can seem a little confusing.

If a function f(x) is continuous at a point xo in its domain, then

$$\lim_{x\to x_o} \; f(x) = f(x_o)$$

which, as we have seen, means

$$ \begin{align} \lim_{x\to x_o^-} \; f(x) &= f(x_o) \; \text{ and} \\ \\ \lim_{x\to x_o^+} \; f(x) &= f(x_o) \end{align}$$

That is, if a function is continuous at a point xo, the limits from right and left must be the same.

Those are properties of a function continuous at a point, but we'd like to have a set of criteria for judging whether a function is continuous over some interval, like [a, b]. There are three criteria that must be met:

It might be more helpful to look at some discontinuities in order to understand function continuity. Here's a made-up function to illustrate the three major kinds:

We can understand these discontinuities very simply at first. They are all discontinuities because if we draw this function from left to right, we have to lift our pen three times, at x = a, b and c. But let's also analyze each in terms of the three continuity criteria.

In the case of a step discontinuity or jump discontinuity like the one at x = a, the left and right limits are different, so the limit does not exist. This step discontinuity violates our third continuity criterion: Because the limits from left and right are different, the limit does not exist.

$$\lim_{x\to x_o^-} \, f(x) \ne \lim_{x\to x_o^+} \, f(x)$$

At the removable discontinuity (x = b), a "hole", the limit exists, but f(b) does not exist. Although we can make x very close to b, f(b) does not exist. This violates our second continuity criterion. The function (not its limit) is not defined at x = b.

At x = c the limit is undefined. From the left the function approaches +∞ and from the right it approaches -∞. None of the continuity criteria are met at x = c.

A function that is differentiable on an interval (a, b) is also continuous on that interval. A function that is continuous on {a, b} is not necessarily differntiable on that interval.

2. Intermediate Value Theorem (IVT)

Intermediate value theorem

If f is a continuous function on the interval [a, b] and C is any number between f(a) and f(b), where f(a) ≠ f(b), then there is a number c in (a, b) such that f(c) = C.

The intermediate value theorem (IVT) simply says that if a continuous function can have two different values, f(a) and f(b), then it must take on every intermediate value between f(a) and f(b) as we move from   a   to   b   along the domain. The function could take on that intermediate value more than once over the interval (a, b).

Think of an airplane landing from an altitude of 10,000 ft. above the runway. If the plane is to reach 0 ft. on the runway, it will have been at every intervening altitude at least once along the way.

If a function has a discontinuity like a hole, it is possible for N not to be in the range of the function over (a, b). That's why we specify that f(x) is continuous in the theorem.

Another way of stating the IVT: A continuous function cannot skip values.

Example 1

Prove that the equation   cos(x) = 0.25   has at least one solution.

Solution: We know that the cosine function is continuous over its domain, (-∞, ∞), and that its range is [-1, 1] over that domain. If we choose an interval over which we think the solution exists, say [0, π/2], we have

cos(0) = 1     and     cos(π/2) = 0

The IVT tells us that because f(x) = cos(x) is continuous, the equation cos(x) = 0.25 must have at least one solution in that interval.

Example 2

Proof of the existence of zeros of a function

We can modify the IVT to prove the existence of zeros simply by making sure that the signs of f(a) and f(b) are opposite: one is above the x-axis and the other below.

Then if we set the value of N at zero, the IVT says that f(x) must equal N at least once, and therefore have at least one zero on (a, b).

3. Extreme Value Theorem

Extreme value theorem

If f is a continuous function on the closed interval [a, b], then f has both an absolute maximum value, f(c), and an absolute minimum value, f(d), for some numbers c and d on [a, b].

The extreme value theorem theorem says that if we consider a closed interval of the domain of a function [a, b] — recall that the square brackets mean that   a   and   b   are included — then the function will have both an absolute maximum and an absolute minimum on that interval. Either the maximum or minimum could be   a   or   b, so when looking for maxima and minima on a closed interval, don't forget to check the endpoints.

Here are four possible cases, first a function with a max and a min between a and b, that is, on (a, b) ← round brackets means endpoints aren't included.

Here's an example in which the maximum is one end of the interval. It may not actually be a max. or min. of the function itself, it just is on this interval.

In this example, the minimum of the function on [a, b] lies on one end (x = b) of the interval

And finally, for a constant function, every point is both a max. and a min.

Why a closed interval?

If we try to express this theorem using an open interval (round brackets), in which the endpoints aren't included, then we can't be sure that the function doesn't grow without bound in either of the ±x directions. Such a function may lack a maximum, a minimum or both.

4. Fermat's Theorem

Fermat's theorem

If a function f has a local minimum or maximum at a point c, and if f'(c) exists, then f'(c) = 0.

Note: The fact that f'(x) = 0 does not necessarily mean that there is a local max. or min. at c. Fermat's theorem does not work in reverse. If there is a max. or min., then the derivative is zero. Existence of a zero derivative is not evidence of a max. or min.

One example of this caveat is the function f(x) = x3, shown below. This function has no maxima or minima, but f'(0) = 0, so the slope of the function is zero at x = 0. This particular point is an inflection point.

Fermat's theorem says that if a function has a local maximum or minimum (which could be global), then the derivative at that point is zero. It's not to difficult to prove Fermat's theorem, so let's do it.

Proof: Referring to the definition of the local maximum above, we see that if a maximum value lies at x = c, then f(x) must be larger than some other value of the function f(x + h), where h can be positive or negative. So:

$$f(c) \ge f(c + h)$$

which rearranges to:

$$f(c + h) - f(c) \le 0$$

We can divide both sides of this inequality by h to make this look like a derivative, then take the limit as h → 0+ from the right:

$$\lim_{h\to 0^+} \; \frac{f(c + h) - f(c)}{h} \le \lim_{h\to 0^+} \; (0)$$

Now we have assumed that f'(c) exists, so the limit from the right must equal the limit in general:

$$\lim_{h\to 0^+} \; \frac{f(c + h) - f(c)}{h} \le 0$$

The expression on the left is f'(c), so we have:

$$f'(c) \le 0$$

Now we could also do the same proof using the limit from the left, which would lead us to the inequality f'(c) ≥ 0, and we'd have the double inequality,

$$0 \le f'(c) \le 0$$

The only way that both inequalities can be true is for f'(c) to be equal to zero, so we have proved Fermat's theorem.

5. Rolle's Theorem

Rolle's theorem

If f is a function that is continuous on the closed interval [a, b] and differentiable on (a, b), and f(a) = f(b), then there is a number c ∈ [a, b] such that f'(c) = 0.

Rolle's theorem simply says that if we take two points of a function having the same y-value, then there must be at least one place on the graph between them where the derivative of the function is zero. You can see this best by looking at the examples below. The first plot shows that there may be more than one zero of the derivative in [a, b].

These to plots show that if f(a) = f(b), there must be at least one critical point like this maximum ...

... or this minimum.

For a constant function, f'(x) = 0 everywhere. The line is also its tangent.

Proof: We take this proof in three parts (refer to the figures above):

  1. f(x) = constant function. If f(x) = c, where c is a constant, then f'(x) = 0 everywhere on (a, b).

  2. f(x) > f(a) for some x in (a, b) [see the two graphs above]. According to the extreme value theorem, f(x) must have a maximum value somewhere on [a, b]. Because f(a) = f(b) and f(x) > f(a), the maximum must be on the open interval (a, b), at some point c. The derivative at c is defined and f(x) is continuous, so the derivative exists and is equal to zero.

  3. f(x) < f(a) for some x in (a, b). According to the extreme value theorem, f(x) has a minimum on [a, b]. Because f(x) < f(a) and f(a) = f(b), then the minimum must be on the open interval (a, b) at some point c. The function is continuous and the derivative exists and is equal to zero.


Show that $f(x) = x^3 + 9x - 4$ has only one real root.

Notice that f(0) = -4 is negative and f(1) = 6 is positive. By the IVT, f(x) has at least one root in [0, 1]. If f(x) had another root - lets call them a and b, then f(a) = f(b) = 0, and Rolle's theorem says that f'(c) = 0

at some point, c, in (a, b). This is not possible because f'(x) = 3x2 + 9, which is always ≥ 9, so f'(c) = 0 has no solutions. We must conclude that the first root, a, is the only real root of f(x).

6. The Mean Value Theorem (MVT)

Mean value theorem

If f is a function that is continuous on the closed interval [a, b] and differentiable on (a, b), and f(a) = f(b), then there is a number c ∈ [a, b] ∋ f'(c) = 0. We write it like this:

$$f'(c) = \frac{f(b) - f(a)}{b - a}$$

(Recall that     means "is an element of" and     means "such that." It's just mathematical shorthand.)

The mean value theorem is a generalized version of Rolle's theorem that takes away the constraint that f(a) = f(b). It says that if f(x) is a continuous function, then somewhere along the curve, the instantaneous value of the slope of f(x) must be the same as the average value of the slope on [a, b]. There may be more than one point on f(x) where f'(x) = f'average.

The mean value theorem says that if I drove an average speed of 42 miles/hour from my house to Boston, at least once during that trip my instantaneous speed must have been 42 mph.

Proof: We employ a geometric proof of the MVT. First take the average value of a function f(x) (dashed line),

$$\text{mean value } \; = \frac{f(b) - f(a)}{b - a}$$

and set a line parallel to that above the function (solid line). If we move that line down until it is tangent to the function, then the slope of that line is the derivative of f(x) at c, f'(c).

These lines are parallel, so they have equal slopes, which proves the MVT.

Now if you're paying attention, you've imagined a situation like the one below, where a line parallel to the mean value brought in from above does not encounter a point of tangency before it encounters either the points (a, f(a)) or (b, f(b)) (which may not have derivatives).

Fair enough, but the solution is easy: Just bring the line in from the bottom like this:

Also notice that in our first figure, there were two points of possible tangency between a and b. That's OK. The MVT guarantees at least one point where the slope of the curve is equal to the mean slope. More than one is OK.

An algebraic proof of the MVT

The equation of a line segment spanning the points (xo, yo) and (x, y), following the usual formula or a line, is:

$$y - y_o = m(x - x_o)$$

If our two points are (a, f(a)) and (b, f(b)), as in this figure

we have:

$$y - f(a) = \frac{f(b) - f(a)}{b - a} (x - a)$$

in which you can easily identify the slope of the segment. We can rearrange to

$$y = \frac{f(b) - f(a)}{b - a} (x - a) + f(a) $$

Now we can employ a trick: Define g(x) = f(x) - y, so

$$g(x) = f(x) - \left[\frac{f(b) - f(a)}{b - a} (x - a) + f(a)\right]$$

Further, notice that g(a) = 0 and g(b) = 0. Here's how. First, if x = a, the second term in brackets is zero:

$$ \begin{align} g(a) &= f(a) - \left[\frac{f(b) - f(a)}{b - a} (a - a) + f(a)\right]\\ \\ &= f(a) - f(a) = 0 \end{align}$$

and if x = b we get a fortuitous cancellation:

$$ \begin{align} g(v) &= f(b) - \left[\frac{f(b) - f(a)}{b - a} (b - a) + f(a)\right]\\ \\ &= f(b) - f(b) + f(a) - f(a) = 0 \end{align}$$

Now our new function g(x) is continuous on [a, b] because f(x) is, and the endpoints, g(a) and g(b) are equal. That means by Rolle's theorem, ∃ c ∈ (a, b) ∋ g(c) = 0. [To refresh your memory on mathematical symbols, ∃ means "there exists"]

Now the straightforward derivative of g(x) is:

$$g'(x) = f'(x) - \frac{f(b) - f(a)}{b - a}$$

Then g(c) is

$$g'(c) = f'(c) - \frac{f(b) - f(a)}{b - a} = 0$$

And finally we derive the mean value theorem:

$$f'(c) = \frac{f(b) - f(a)}{b - a}$$

The mean value theorem is the basis of much of calculus.

1. Proof of "intuitive" concepts

Up until now we've made and accepted three crucial statements, more or less because they seem fine. They are summarized below.

Now we are in a position to prove these propositions. We begin with the MVT in the form

$$f(b) - f(a) = f'(c)(b - a)$$

If we assume that b > a (as in the three graphs above), then (b - a) is always positive. Now it's easy to see that

  1. If f'(c) < 0, then f(a) > f(b), which means that the function is decreasing

  2. If f'(c) > 0, then f(a) < f(b), which means that the function is increasing, and

  3. If f'(c) = 0, then f(a) = f(b) and the function is constant over (a, b)

2. Simplification of some calculations

The mean value theorem (MVT) guarantees the existence of the mean value somewhere along the function, but it doesn't tell us exactly where it is. What the MVT does is replace a difference between function values, f(b) - f(a) with the derivative of the function, f'(c) times a simple difference between coordinates (b - a):

$$f(b) - f(a) = f'(c)(b - a)$$

For example, to compute the value of cos(1.8) - cos(2.9) is more difficult than to simply multiply (1.8 - 2.9) by a derivative.

3. A necessary component of developing integral calculus

In other sections, you'll see how each time we develop the integral for a specific case, rectilinear vs. polar coordinates, for example, we use the MVT.

You can find it used in developing the definite integral and the integrals of polar functions.

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