Mean Value Theorem

Question

Does the function satisfy the hypotheses of the Mean Value Theorem on the given interval?

f(x) = ln(x),    [1, 5]

Yes, it does not matter if f is continuous or differentiable, every function satisfies the Mean Value Theorem.

Yes, f is continuous on [1, 5] and differentiable on (1, 5).    

No, f is not continuous on [1, 5].

No, f is continuous on [1, 5] but not differentiable on (1, 5).

There is not enough information to verify if this function satisfies the Mean Value Theorem.

I believe that perhaps it is not continuous at [1,5], and is not differentiable on (1,5) due to the fact that we are trying to get the answer equal to 0.

If it satisfies the hypotheses, find all numbers c that satisfy the conclusion of the Mean Value Theorem. I

As of this moment, I am very confused abuout the mean theoram value, and would like to ask for help on this question on wyzant

Answer 1

To answer this question, we first must consider what the mean value theorem (MVT) is about. The MVT states that if f is a function that satisfies these two hypotheses i) f is continuous on the closed interval [a,b] and ii) f is differentiable on the closed interval (a,b), then there is a number c in the interval (a,b) such that f'(c) = f(b)-f(a)/(b-a). In this case, f(x) = ln(x) is continuous on the interval [1,5] and differentiable on (1,5) (we can graph and check this). Also, the derivative of f(x) is 1/x. Therefore, by the MVT, there exists a value, c, which equals f'(c) = 1/c = f(5)-f(1)/(5-1) --> ln(5)-ln(1)=(5-1). Upon simplifying, the point c is 4/ln(5).

Hope this helps!

Answer 2

The function f(x) = ln(x) does satisfy the conditions for MVT because it is continuous on [1,5] and differentiable on (1,5).

How to apply MVT? Well, we want to find the value c such that

f'(c) = (f(5)-f(1))/(5-1) = (ln(5)-ln(1))/4 = ln(5)/4. So, we set

f'(c) = 1/c = ln(5)/4, so that c = 4/ln(5) = 2.485.