Question

What value of x satisfies the conclusion of the mean value theorem for f(x) = ln(x3) over the interval [1, e2]?

Answer 1

`x \approx 3.195` satisfies the conclusion of the Mean Value Theorem for
`f(x) = \ln x^(3)` over the interval
`[1,e^(2)]`.

Explanation:

According to the Mean Value Theorem, for all function that is differentiable over the interval
`[a, b]`, there is at a value
`c` within the interval such that:

`f'(c) = (f(b)-f(a))/(b-a)` (1)

Where:

`a`,
`b` - Lower and upper bounds.

`f(a)`,
`f(b)` - Function evaluated at lower and upper bounds.

`f'(c)` - First derivative of the function evaluated at
`c`.

If we know that
`f(x) = \ln x^(3) = 3\cdot \ln x`,
`f'(x) = (3)/(x)`,
`a = 1` and
`b = e^(2)`, then we find that:

`(3)/(c) = (3\cdot \ln e^(2)-3\cdot \ln 1)/(e^(2)-1)`

`(3)/(c) = (6\cdot \ln e-3\cdot \ln 1 )/(e^(2)-1 )`

`(3)/(c) = (6)/(e^(2)-1)`

`c = (1)/(2)\cdot (e^(2)-1)`

`c \approx 3.195`

`x \approx 3.195` satisfies the conclusion of the Mean Value Theorem for
`f(x) = \ln x^(3)` over the interval
`[1,e^(2)]`.