# How do you find the domain of f(x) =1/sqrt(-x^2+5x-4)?

Aug 25, 2017

The domain is $x \in \left(1 , 4\right)$

#### Explanation:

For the square root and the division, the conditions are

$\left(- {x}^{2} + 5 x - 4\right) > 0$ and $x \ne 1$ and $x \ne 4$

Let $f \left(x\right) = \left(- {x}^{2} + 5 x - 4\right) = - \left({x}^{2} - 5 x + 4\right) = - \left(x - 1\right) \left(x - 4\right)$

We can build the sign chart

$\textcolor{w h i t e}{a a a a}$$x$$\textcolor{w h i t e}{a a a a a a a}$$- \infty$$\textcolor{w h i t e}{a a a a a a}$$1$$\textcolor{w h i t e}{a a a a a a a a}$$4$$\textcolor{w h i t e}{a a a a a a}$$+ \infty$

$\textcolor{w h i t e}{a a a a}$$- \left(x - 1\right)$$\textcolor{w h i t e}{a a a a a}$$+$$\textcolor{w h i t e}{a a a}$$| |$$\textcolor{w h i t e}{a a a}$$-$$\textcolor{w h i t e}{a a a a a a}$$-$

$\textcolor{w h i t e}{a a a a}$$\left(x - 4\right)$$\textcolor{w h i t e}{a a a a a a a}$$-$$\textcolor{w h i t e}{a a a}$color(white)(aaaa)-$\textcolor{w h i t e}{a a}$$| |$$\textcolor{w h i t e}{a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$f \left(x\right)$$\textcolor{w h i t e}{a a a a a a a a a a}$$-$$\textcolor{w h i t e}{a a}$$| |$$\textcolor{w h i t e}{a a a}$$+$$\textcolor{w h i t e}{a a}$$| |$$\textcolor{w h i t e}{a a a}$$-$

Therefore,

$f \left(x\right) > 0$, when $x \in \left(1 , 4\right)$

graph{1/sqrt(-x^2+5x-4) [-9.55, 18.93, -4.48, 9.76]}