16. Define by . Show that f satisfies the conditions of Rolle's theorem and find c such that f'(c) = 0.
Well, what are the hypothsis of Rolle's Theorem.
f is cont on [0,2] and differentiable on (0,2).
Lastly $\displaystyle f(0)=\sqrt{2(0)-0^2}=0,f(2)=\sqrt{2(2)-(2)^2}=0$
So there exits a $\displaystyle c \in (0,2)$ such that$\displaystyle f'(c)=0$
taking the derivative we get
$\displaystyle \frac{df}{dx}=\frac{1}{2}\cdot \frac{2-2x}{\sqrt{2x-x^2}}=\frac{1-x}{\sqrt{2x-x^2}}$
setting equal to zero and solveing gives c=1 or
$\displaystyle f'(c)=0 \mbox{ when } c=1$