Answer
The graph of the function has at least two x-intercepts in the interval $(0,2)$.
Work Step by Step
$y = x^2-3+\frac{1}{x}$
We can see that this function is continuous in the interval $(0,2)$.
Let $x = \frac{1}{3}$
$y = (\frac{1}{3})^2-3+\frac{1}{1/3} = \frac{1}{9}$
Let $x = 1$
$y = (1)^2-3+\frac{1}{1} = -1$
Let $x = \frac{5}{3}$
$y = (\frac{5}{3})^2-3+\frac{1}{5/3} = \frac{17}{45}$
By the Intermediate Value Theorem, there is a number $b$ in $(\frac{1}{3}, 1)$ such that $b^2-3+\frac{1}{b} = 0$, and there is a number $c$ in $(1,\frac{5}{3})$ such that $c^2-3+\frac{1}{c} = 0$
The graph of the function has at least two x-intercepts in the interval $(0,2)$
