Question

find the domain of the function. (enter your answer in interval notation.)
$h(x) = \frac{1}{\sqrt4{x^2 - 7x}}$

Explanation:

Step1: Analyze the denominator (radical and zero)

For the function \( h(x)=\frac{1}{\sqrt[4]{x^{2}-7x}} \), the denominator \(\sqrt[4]{x^{2}-7x}\) must satisfy two conditions: the radicand \(x^{2}-7x\) must be positive (since we can't have zero in the denominator for a fourth - root in the denominator, and the fourth - root of a non - positive number is not real in the set of real numbers when the radicand is negative) and the radicand must be defined (for real numbers, the radicand of an even - root must be non - negative, but since we can't have zero in the denominator, we need \(x^{2}-7x>0\)).

Factor the quadratic expression: \(x^{2}-7x=x(x - 7)\).

Step2: Solve the inequality \(x(x - 7)>0\)

To solve the inequality \(y = x(x - 7)>0\), we first find the roots of the equation \(x(x - 7)=0\), which are \(x = 0\) and \(x=7\).

These roots divide the real number line into three intervals: \((-\infty,0)\), \((0,7)\), and \((7,\infty)\).

• For the interval \((-\infty,0)\): Let's take a test point, say \(x=-1\). Then \(y=(-1)\times(-1 - 7)=(-1)\times(-8) = 8>0\). So the inequality is satisfied on \((-\infty,0)\).
• For the interval \((0,7)\): Let's take a test point, say \(x = 1\). Then \(y=(1)\times(1 - 7)=(1)\times(-6)=-6<0\). So the inequality is not satisfied on \((0,7)\).
• For the interval \((7,\infty)\): Let's take a test point, say \(x = 8\). Then \(y=(8)\times(8 - 7)=(8)\times(1)=8>0\). So the inequality is satisfied on \((7,\infty)\).

Answer:

\((-\infty,0)\cup(7,\infty)\)