The interval in question is `[a,b] = [1,4]`

The mean value theorem says that if a function is continuous on `[a,b]` (closed interval) and differentiable on `(a,b)` (open interval) then there is a point `c` in `(a,b)` such that

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

Now, we want `f'(c) = (ln(b) -ln(a))/(b-a) = ln(b/a)/(b-a) = ln(4)/3 = ln(4^(1/3))`

Using `ln(b) - ln(a) = ln(b/a)` and `rlns = ln(s^r)`

We have `f(x) = ln x`

So we want `c` such that `lnc = ln(4^(1/3))`

This implies `c = 4^(1/3) = 1.587`

**Yes, c = 1.587**

Posted on

## We’ll help your grades soar

Start your 48-hour free trial and unlock all the summaries, Q&A, and analyses you need to get better grades now.

- 30,000+ book summaries
- 20% study tools discount
- Ad-free content
- PDF downloads
- 300,000+ answers
- 5-star customer support

Already a member? Log in here.

Are you a teacher? Sign up now