Question

G verify that the mean value theorem applies to f(x)= x+2/x on the interval [1,2] find point c satisfying conclusion

Answer 1

The mean value theorem requires that a function
`f(x)` is continuous on the interval
`[a,b]` and differentiable on the same open interval
`(a,b)`.

Since the only point of discontinuity of your function is
`x = 0`, where the denominator of 2/x equals zero, the function is defined, continuous and differentiable on the interval
`[1,2]`.

Since the hypothesis of the mean value theorem are verified, we can claim that there exists a point
`c \in (a,b)` such that

`f'(c) = \cfrac{f(b)-f(a)}{b-a}`

In your case,
`b = 2, a = 1`, so the expression becomes

`f'(c) = \cfrac{f(2)-f(1)}{2-1} = \cfrac{3-3}{1} = 0`

Let's compute the derivative:

`f(x) = x+\cfrac{2}{x} \implies f'(x) = 1-\cfrac{2}{x^2}`

So, we're looking for a point c such that

`f'(c)=0 \implies 1-\cfrac{2}{c^2} = 0 \iff 1 = \cfrac{2}{c^2} \iff c^2=2 \iff c = \pm√(2)`

Since we are looking for a point in
`(1,2)`, we accept the solution
`c = √(2)`