Question

39-46 find the domain of the function.

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

Short answer

The domain of the function is \(\left( { - \infty , - 1} \right) \cup \left( {5,\infty } \right)\).

Step-by-step solution

Define the domain of the function

Let \(D\) and \(E\) be the sets of real numbers. Set \(D\) is referred to as thedomainof the function. The domain of the function is the set of all inputs for which the formula produces a real number output.

Determine the domain of the function

The function \(h\left( x \right) = \sqrt {{x^2} - 4x - 5} \) is defined if \({x^2} - 4x - 5 \ge 0\). Then

\({x^2} - 4x - 5 \ge 0\)

\(\left( {x + 1} \right)\left( {x - 5} \right) \ge 0.\)

The polynomial \(p\left( x \right) = {x^2} - 4x - 5\) can only change signs at its zeros.

Test the values of \(x\) on the interval separated by \(x = - 1\) and \(x = 5\), as shown below:

\(p\left( { - 2} \right) = 4 + 8 - 5\)

\( = 7 > 0\)

\(p\left( 0 \right) = 0 - 0 - 5\)

\( = - 5 < 0\)

The domain of \(h\) corresponds to the solution intervals of \(p\left( x \right) \ge 0\) as \(\left\{ {x|x \le - 1\,\,{\mathop{\rm or}\nolimits} \,\,x \ge 5} \right\} = \left( { - \infty , - 1} \right) \cup \left( {5,\infty } \right)\).

Thus, the domain of the function is \(\left( { - \infty , - 1} \right) \cup \left( {5,\infty } \right)\).