Question

Find the domain of the function

\(h\left( x \right) = \frac{1}{{\sqrt(4){{{x^2} - 5x}}}}\)

Short answer

The domain of the function \(h\left( x \right) = \frac{1}{{\sqrt(4){{{x^2} - 5x}}}}\)is\(\left( { - \infty ,\;0} \right) \cup \left( {5,\;\infty } \right)\).

Step-by-step solution

Describe the domain

The domain is the set of all values at which the given function is defined.

Find the domain of the function \(h\left( x \right) = \frac{1}{{\sqrt(4){{{x^2} - 5x}}}}\)

The fourth root exists only for positive quantities and the denominator should not be equal to\(0\).

\(\begin{array}{c}{x^2} - 5x > 0\\x\left( {x - 5} \right) > 0\end{array}\)

Let\(f\left( x \right) = x\left( {x - 5} \right)\).

If \(f\left( x \right) = 0\) then roots of the function are \(0\)and\(5\).

Construct the following table.

Interval	Sign of \(f\left( x \right)\)
\(\left( { - \infty ,\;0} \right)\)	+
\(\left( {0,\;5} \right)\)	-
\(\left( {5,\;\infty } \right)\)	+

From the above table, it can be observed that \(x\left( {x - 5} \right) > 0\) satisfies when \(x \in \left( { - \infty ,\;0} \right)\)or\(x \in \left( {5,\;\infty } \right)\).

Therefore, the domain of the function \(h\left( x \right) = \frac{1}{{\sqrt(4){{{x^2} - 5x}}}}\)is\(\left( { - \infty ,\;0} \right) \cup \left( {5,\;\infty } \right)\).