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.

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.

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.

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.

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).

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.

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.

... 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.

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).

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.

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:

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

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)

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

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.

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