Question

How do you determine all values of c that satisfy the mean value theorem on the interval [6,10] for `f(x)= ln (x-5)`?

Answer 1

`c= (4 + ln3125)/ln5`

Explanation:

The function `ln(x- 5)` is defined and continuous for all `x > 5`.

We can therefore most certainly use the mean value theorem to solve for `C`. The theorem states that if `f(x)` is defined, continuous and differentiable on the interval `[a, b]`, then there is certainly one number `C` that satisfies `f'(C) = (f(b) - f(a))/( b - a)`. In other words, the average rate of change will be equal to the instantaneous rate of change at some point `C` on the function.

We start by finding the derivative of `f(x)`. Let `y = lnu` and `u = x - 5`.

Then `dy/(du) = 1/u` and `(du)/dx = 1`.

`dy/dx = 1/u xx 1`

`dy/dx= 1/(x - 5)`

By the mean value theorem:

`1/(c- 5) = (f(10) - f(6))/(10 - 6)`

`1/(c- 5) =(ln5 - ln1)/4`

`1/(c- 5) = (ln5 - 0)/4`

`4 = (c- 5)ln5`

`4 = cln5 - 5ln5`

`4 + ln3125 = cln5`

`c= (4 + ln3125)/ln5 ~= 7.485`

Hopefully this helps!