Find the Domain f(x)=(x-5)/( square root of x)

$f (x) = \frac{x - 5}{\sqrt{x}}$

Step 1

Set the radicand in $\sqrt{x}$ greater than or equal to $0$ to find where the expression is defined.

$x \geq 0$

Step 2

Set the denominator in $\frac{x - 5}{\sqrt{x}}$ equal to $0$ to find where the expression is undefined.

$\sqrt{x} = 0$

Step 3

Solve for $x$ .

To remove the radical on the left side of the equation, square both sides of the equation.

${\sqrt{x}}^{2} = 0^{2}$

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{x}$ as $x^{\frac{1}{2}}$ .

${(x^{\frac{1}{2}})}^{2} = 0^{2}$

Simplify the left side.

Simplify ${(x^{\frac{1}{2}})}^{2}$ .

Multiply the exponents in ${(x^{\frac{1}{2}})}^{2}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$x^{\frac{1}{2} \cdot 2} = 0^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

$x^{\frac{1}{2} \cdot 2} = 0^{2}$

Rewrite the expression.

$x^{1} = 0^{2}$

Simplify.

$x = 0^{2}$

Simplify the right side.

Raising $0$ to any positive power yields $0$ .

$x = 0$

Step 4

The domain is all values of $x$ that make the expression defined.

Interval Notation:

$(0, \infty)$

Set-Builder Notation:

${x | x > 0}$

Step 5