![\bf f(x)=x+\cfrac{1}{x}\qquad \left[\frac{1}{2},2 \right]\\\\ -----------------------------\\\\ \cfrac{df}{dx}=1+\left(-1x^{-2} \right)\implies \cfrac{df}{dx}=1-\cfrac{1}{x^2} \\\\\\ f'(c)=1-\cfrac{1}{c^2}\quad \quad 1-\cfrac{1}{c^2}=\cfrac{f(2)-f\left( \frac{1}{2} \right)}{2-\frac{1}{2}} \\\\\\ 1-\cfrac{1}{c^2}=\cfrac{\frac{5}{2}-\frac{5}{2}}{\frac{3}{2}}\implies 1-\cfrac{1}{c^2}=\cfrac{0}{\frac{3}{2}}\implies 1-\cfrac{1}{c^2}=0 \\\\\\ 1=\cfrac{1}{c^2}\implies c^2=1\implies c=\pm\sqrt{1}\implies c=\pm 1](/tpl/images/0251/9847/a3c9b.png)
Checking the mean value theorem:
number 3
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