Question

How do I find the numbers `c` that satisfy the Mean Value Theorem for `f(x)=x/(x+2)` on the interval `[1,4]` ?

Answer 1

The mean value theorem guarantees that there exists a number `c` in `(1,4)` such that
`f'(c)={f(4)-f(1)}/{4-1}`.
The actual value of `c` is `-2+3sqrt{2}`.

Let us find the left-hand side of the above equation,
By Quotient Rule,
`f'(x)={1cdot(x+2)-xcdot1}/{(x+2)^2}=2/(x+2)^2`
`Rightarrow f'(c)=2/(c+2)^2`

Let us find the right-hand side,
`{f(4)-f(1)}/{4-1}={4/6-1/3}/{3}=1/9`

By setting the left-hand side and the right-hand side equal to each other,
`2/(c+2)^2=1/9`

by taking the reciprocal,
`(c+2)^2/2=9`

by multiplying by 2,
`(c+2)^2=18`

by taking the square-root,
`c+2=pm sqrt{18}=pm3sqrt{2}`

by subtracting 2,
`c=-2 pm3sqrt{2}`

Since `1 < c < 4`,
`c=-2+3sqrt{2}`