Question

Given the function `f(x)= ln x^2`, how do you determine whether f satisfies the hypotheses of the Mean Value Theorem on the interval [1,3] and find the c?

Answer 1

`c = 2/ln(3)`

Explanation:

First of all, let's remind that the Mean Value Theorem states that, if `f` is a continuous function on the closed interval `[a,b]`, and differentiable on the open interval `(a,b)`, then there exists some `c` in `(a,b)` such that

`f'(c) = (f(b)-f(a))/(b-a)`

id est, there exists an inner point in which the tangent line is parallel to the line connecting `f(a)` and `f(b)`.

Since your function is continuous in `[1,3]` and differentiable in `(1,3)`, we can invoke the theorem.

Moreover, we can observe that `ln(x^2)=2ln(x)`, which will simplify a bit our computations. In fact, we have:

• `f'(x) = 2/x`
• `f(b) = f(3) = 2ln(3)`
• `f(a) = f(1) = 2ln(1)=0`
• `b-a = 3-1 = 2`

So, we need to solve for `c` the equation

`f'(c) = (f(b)-f(a))/(b-a) \to 2/c = (2ln(3))/2 = ln(3)`

Inverting both sides, we have

`2/c = ln(3) \iff c/2 = 1/ln(3)`

And finally, multiplying by `2` both sides,

`c/2 = 1/ln(3) \iff c = 2/ln(3)`

We can also check that `2/ln(3)` is approximately `1.8204...`, so it is indeed a number between `1` and `3`.