Question

Checking the Mean Value Theorem:Number 3

Answer 1

$\bf f(x)=x+\cfrac{1}{x}\qquad \left[\frac{1}{2},2 \right]\\\\ -----------------------------\\\\ \cfrac{df}{dx}=1+\left(-1x^{-2} \right)\implies \cfrac{df}{dx}=1-\cfrac{1}{x^2} \\\\\\ f'(c)=1-\cfrac{1}{c^2}\quad \quad 1-\cfrac{1}{c^2}=\cfrac{f(2)-f\left( \frac{1}{2} \right)}{2-\frac{1}{2}} \\\\\\ 1-\cfrac{1}{c^2}=\cfrac{\frac{5}{2}-\frac{5}{2}}{\frac{3}{2}}\implies 1-\cfrac{1}{c^2}=\cfrac{0}{\frac{3}{2}}\implies 1-\cfrac{1}{c^2}=0 \\\\\\ 1=\cfrac{1}{c^2}\implies c^2=1\implies c=\pm\sqrt{1}\implies c=\pm 1$

there's a quick graph below of the bounds and the tangent at "c"

not happening -2 or 2 will have a tangent parallel to a,b, needless to say -2 is out of the range [a,b] anyway, so the only value is really 1, on the positive 1st quadrant