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.
|(not really an existence theorem, but an important foundation for the others)|
|A continuous function cannot skip values.|
|Any continuous function has a global max. & min. on a closed interval.|
|f(x) = 0 at a maximum or minimum.|
|Special case of the mean value theorem.|
|Any continuous function takes on its average value at least once in an open interval.|
If a function f(x) is continuous at a point xo in its domain, then
which, as we have seen, means
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.
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.
Another way of stating the IVT: A continuous function cannot skip values.
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.
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).
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.
If we try to express this theorem using an open interval, on 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.
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:
which rearranges to:
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:
Now we have assumed that f'(c) exists, so the limit from the right must equal the limit in general:
The expression on the left is f'(c), so we have:
Now we could also do the same proof using the limit from the left, which would lead us to the inequality f'(c) ≥ 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.
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 in the box above):
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).
(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) = faverage.
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),
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.
If our two points are (a, f(a)) and (b, f(b)), as in this figure
in which you can easily identify the slope of the segment. We can rearrange to
Now we can employ a trick: Define g(x) = f(x) - y, so
Further, notice that g(a) = 0 and g(b) = 0. Here's how. First, if x = a, the second term in brackets is zero:
and if x = b we get a fortuitous cancellation:
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:
Then g(c) is
And finally we derive the mean value theorem:
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
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
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):
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.
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