# How do I find the numbers c that satisfy Rolle's Theorem for f(x)=sqrt(x)-x/3 on the interval [0,9] ?

Sep 22, 2014

$c = \frac{9}{4}$

Let us look at some details.

By taking the derivative,

$f ' \left(x\right) = \frac{1}{2 \sqrt{x}} - \frac{1}{3}$

By solving $f ' \left(c\right) = 0$,

$f ' \left(c\right) = \frac{1}{2 \sqrt{c}} - \frac{1}{3} = 0$

by adding $\frac{1}{3}$,

$R i g h t a r r o w \frac{1}{2 \sqrt{c}} = \frac{1}{3}$

by taking the reciprocal,

$R i g h t a r r o w 2 \sqrt{c} = 3$

by dividing by 2,

$R i g h t a r r o w \sqrt{c} = \frac{3}{2}$

by squaring,

$R i g h t a r r o w c = \frac{9}{4}$