Answer
$f(2)≈ 0.414$,
$f(3)≈ −0.601.$
There is a root of the equation $\frac{2}{3} − x + \sqrt x= 0,$ or
$\frac{2}{x} = x − \sqrt x,$ in the interval $(2, 3)$.
Work Step by Step
Given:
$\frac{2}{x} = x − \sqrt x$.
The given equation is equivalent to the equation $\frac{2}{x}- x + \sqrt x = 0$. $f(x)=\frac{2}{x} = x − \sqrt x$ is continuous on the
interval [2, 3],
$f(2) = \frac{2}{2} − 2 + \sqrt 2 $
$= 1 − 2 + \sqrt 2 $
$=-1+1.414$
$f(2)≈ 0.414$,
and
$f(3) =\frac{ 2}{3} -3 + \sqrt 3$
$=\frac{2-9+3\sqrt 3}{3}$
$=\frac{-7+3(1.732)}{3}$
$=\frac{-7+5.196}{3}$
$=\frac{-1.804}{3}$
$f(3)≈ −0.601.$
Since $f(2) > 0 >f(3),$ there is a number $c$
in $(2, 3)$ such that $f(c)=0$ by the Intermediate Value Theorem. Thus, there is a root of the equation $\frac{2}{3} − x + \sqrt x= 0,$ or
$\frac{2}{x} = x − \sqrt x,$ in the interval $(2, 3)$.
