Question

Why is the function not differentiable?

Consider the function f(x) = 3(x - 2)^2/5
A)The function is not differentiable at x = 2. Why not?
B)Will the Mean Value Theorem apply in the interval [2, 5] even though the function is not differentiable at x = 2? Explain.
C) Will the Mean Value Theorem apply to the interval [1, 4]? Explain.

Answer 1

`A)` The derivative doesn't exist
`B)` Yes
`C)` No

Explanation:

Question A
You can see this multiple different ways. Either we can differentiate the function to find:
`f'(x)=6/5(x-2)^(-3/5)=6/(5(x-2)^(3/5))`
which is undefined at `x=2`.

Or, we can look at the limit:
`lim_(h->0)(f(2+h)-f(2))/h=lim_(h->0)(3(2+h-2)^(2/5)-3(2-2)^(3/5))/h=`
`=lim_(h->0)0/h`
This limit limit doesn't exist, which means that the derivative doesn't exist in that point.

Question B
Yes, the Mean Value Theorem does apply. The differentiability condition in the Mean Value Theorem only requires the function to be differentiable on the open interval `(a,b)` (IE not `a` and `b` themselves), so on the interval `[2,5]`, the theorem applies because the function is differentiable on the open interval `(2,5)`.

We can also see that there is indeed a point with the average slope in that interval:

Question C
No. As previously mentioned, the Mean Value Theorem requires the function to be entirely differentiable on the open interval `(1,4)`, and we previously mentioned that the function isn't differentiable at `x=2`, which lies in that interval. This means that the function isn't differentiable on the interval, and therefor the Mean Value Theorem doesn't apply.

We can also see that there is no point in the interval which contains the average slope on this function, because of the "sharp bend" in the curve.