Question

How do you determine all values of c that satisfy the mean value theorem on the interval [2,5] for `f(x) = 1 / (x-1)`?

Answer 1

`AA c \in [f(5), f(2)] = [1/4, 1]`, `EE y \in [2, 5]` such that `f(y) = c`.
Explanation below.

Explanation:

Since `f` is a quotient of continuous function :
`f(x) = (g(x))/(h(x)), g(x) = 1` and `h(x) = x-1`,
it is defined and continuous for all `x \in RR \backslash {1}` (`f(1)` is not defined because we would divide by zero).

Therefore, the mean value theorem is satisfied for all closed interval `[a, b] \subset RR \backslash {1}`.

Since `[2, 5] \subset RR \backslash {1}` and `[2, 5]` is closed, we have, by the mean value theorem, that `AA c \in [f(5), f(2)] = [1/4, 1]`, `EE y \in [2, 5]` such that `f(y) = c`.

Answer 2

`c=3 or c =-1`

Explanation:

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

`f'(c)=(f(5)-f(2))/(5-2)`

`-1/(c-1)^2 = (1/4 -1)/3 =(-3/4)/3 = -3/4 *1/3=-1/4`

`-1=-1/4 (c-1)^2`

`4 = c^2-2c+1`

`0=c^2-2c-3`

`(c-3)(c+1)=0`

`c=3 or c =-1`