Rolle's Theorem states:
If f is continuous on [a,b] and differentiable on (a,b), and if f(a)=f(b)=0, then there is some c in the interval (a,b) such that f'(c)=0.
But what does this mean?
This packet will help you think graphically about the meaning of this important Calculus theorem.
Make yourself a set of x-y axes and label two points on the x-axis as A and B, as below:
Now sketch a nice, smooth function that goes through those two points. It doesn't matter what else happens-whether your function is a polynomial, or periodic, whatever. It just needs to be (1) smooth, and (2) continuous. For example:
Can you find a place between A and B where the derivative equals zero, as below?
Of course you can! Rolle's Theorem states that it is always possible to find such a point under these conditions.
Let’s look at a concrete example. Let a=3 and b=7. Then any smooth, continuous function that goes through the points (3,0) and (7,0) will have to have some point in the interval (3,7) where f’(c)=0.
One such f is f(x)=(x-3)(x-7). This function is equal to zero at x=3 and x=7. Furthermore, it is a polynomial so it satisfies the hypothesis that f is continuous and differentiable. Rolle’s Theorem tells us that there must be a point in the interval (3,7) where f’=0.
From the picture, it appears that this point is at x=5.
We can check it: f(x)=x^{2}-10x+21, so f’(x)=2x-10. Then f’(5)=2*5-10=0.
It is important to note, though, that Rolle’s Theorem does not tell us how to find the point where f’(x)=0; it only tells us that this point exists.
Another function that has these same roots is f(x)=2^{x}-(30x-82). The derivative of this function is very hard to work with (and to solve for zero!), but that derivative does in fact exist and Rolle’s Theorem tells us that there is a solution: f’(x)=0 somewhere in the interval (3,7).
But in this instance, we don't know exactly where it is. We can estimate from the graph, but Rolle's Theorem doesn't give us a formula for finding the point in question. Instead it guarantees that such a point exists.
So what is Rolle's Theorem good for, anyway?
It is the basis for proving the Mean Value Theorem, which turns out to be really important for doing lots of stuff in Calculus. On a simplistic and practical level, a standard example of the Mean Value Theorem is the following:
If you drive from Saint Paul to Duluth, a distance of 150 miles, Google Maps tells us that it will take 2.5 hours, an average of 60 miles per hour. But the speed limit on much of I-35 along the way is 70 mph. And there will be some city driving where we drive more slowly. Intuitively, we know that if we average 60 mph, we must at some point drive exactly 60 mph. The Mean Value Theorem formalizes this and assures us that our intuition is correct.
Of course that's not what we really need the Mean Value Theorem for-we need it for more abstract mathematical work. But it's always nice to know when formal mathematics matches with our informal intuitions.
verify roll e's theorem for the function: f(x)=(x-2)^6(x-5)^11 in the interval, (2<x<5)
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