## Mean Value Theorem

Today’s question (problem submitted by a reader) covers the mean value theorem from Calc. I. This one is a bit subtle in the sense that it raises visibility to the existence of a derivative in an interval, even when the function is continuous. The problem was find the point, c, in the interval [-8,8] for the function f(x) = x^{2/3} that satisifes the mean value theorem.

For review, the mean value theorem states that if a function, f(x), is continuous in [a,b] and differentiable in (a,b), then there exists at least one point, c, in (a,b) at which f(b) – f(a) = f'(c)(b-a). Geometrically, the mean value theroem implies that there is a point in the specified interval at which the slope of the tangent at that point is equal to the slope of the chord between the endpoints of the interval. In terms of velocity, one might say that if a trip over some time interval averaged 60mph, then there must exist at least one point during the trip at which the instantaneous velocity was exactly 60mph.

For the problem at hand, f(b)-f(a)/b-a = 0/16 = 0, so the slope of the chord between the endpoints is zero. The derivative, f'(x) is (2/3)x^{-1/3}, which exists everywhere in (-8,8) except zero. Theoretically, we could just stop here. The derivative is negative in [-8,0) and positive in (0,8]. It is never **exactly** zero in the interval (-8,8). Like the spoon in The Matrix, there is no point satisfying the mean value theorem in this problem.

Now, it is possible that a function may not be differentiable in an interval (a,b) and still have one or more points satisfying the conditions of the MVT. However, one can not use the MVT to guarantee the existence of such a point(s). In this example, it was game over as soon as you showed the function was not differentiable in the interval.

## Limit Proofs

At first, I thought someone was sending me their calculus homework, but this time of the semester, it’s most likely a question that was asked on a final. Anyway, I don’t mind posting solutions on topics that are of general interest to high-school and college students … whenever I have the time 🙂

Question: What is the limit as x→2 of the function x^{2} + 3 , using only the definition of a limit?

So-called epsilon-delta proofs can be cumbersome for polynomials of degree greater than one because of the apparent lack of a single delta value. To review, limit as x→c of f(x) = f(c) means that for any ε > 0, there exists δ > 0 such that |x – c| < δ ==> |f(c) – f(x)| < ε . Intuitively, the definition means we can make f(x) arbitrarily close to f(c) by making x ‘sufficiently’ close to c.

For the problem at hand, it appears that the limit as x→2 is 7. To prove this assertion using the definition, then given any ε > 0 , we must show there exists a δ > 0 (remember that the delta value is a function of epsilon) such that |f(x) – 7| < ε. Now,

|f(x) – 7| < ε ==> |x^{2} – 4| = | (x+2)(x-2) | = |x+2||x-2| < ε . Notice that lim as x→2 of |x-2| = 0 (this is easily proven from the definition), so the trick is to choose an ‘easy’ bound for one factor and limit the other. Consider two deltas, δ_{1} and δ_{2}. Since x-2 approaches zero as x→2, set δ_{1} = 1. Now,

|x-2| < δ_{1} = 1 ==> -1 < x – 2 < 1 ==> 3 < x + 2 < 5 ==> |x+2| < 5. Take δ_{2} = ε /5 and δ = min(δ_{1}, δ_{2}).

Case 1: Any ε ≥ 5. |x-2| < δ = min(δ_{1}, δ_{2}) ==> |x-2| < 1, so

|f(x)-7| = |x+2||x-2| < 5|x-2| < 5, which is less than ε .

Case 2: Any 0 < ε < 5. |x-2| < δ = min(δ_{1}, δ_{2}) ==> |x-2| < ε/5, so

|f(x)-7| = |x+2||x-2| < 5|x-2| < 5(ε/5) < ε .

Strictly speaking, separate cases were not necessary, but I find that students understand the concept easier the first few times this way.