Answer:
[tex]k(h(x)) = \sqrt{x^2 - 3x - 2}[/tex]
[tex]x = (\infty, -0.56]\ u\ [3.56, \infty)[/tex]
Step-by-step explanation:
Given
[tex]h(x) = x^2 - 3x[/tex]
[tex]k(x) = \sqrt{x - 2}[/tex]
Solving (a): k(h(x))
[tex]k(x) = \sqrt{x - 2}[/tex]
Replace x with h(x)
[tex]k(h(x)) = \sqrt{h(x) - 2}[/tex]
Substitute:[tex]h(x) = x^2 - 3x[/tex]
[tex]k(h(x)) = \sqrt{x^2 - 3x - 2}[/tex]
Solving (b): The domain
For the function to be defined, the expression in the root must be greater than or equal to 0; i.e.
[tex]x^2 - 3x - 2 \ge 0[/tex]
Solve for x;
Using a calculator, we have:
[tex]x \le -0.56[/tex] and [tex]x \ge 3.56[/tex] --- approximated
So, the domain is:
[tex]x = (\infty, -0.56]\ u\ [3.56, \infty)[/tex]
