Question

Checking the Mean Value Theorem:

Number 3

Answer 1

`\bf f(x)=x+\cfrac{1}{x}\qquad \left[(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( (1)/(2) \right)}{2-(1)/(2)} \\\\\\ 1-\cfrac{1}{c^2}=\cfrac{(5)/(2)-(5)/(2)}{(3)/(2)}\implies 1-\cfrac{1}{c^2}=\cfrac{0}{(3)/(2)}\implies 1-\cfrac{1}{c^2}=0 \\\\\\ 1=\cfrac{1}{c^2}\implies c^2=1\implies c=\pm√(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