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

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

Since we are looking for a point in $(1,2)$, we accept the solution $c = \sqrt{2}$