Checkpoint: Combinations of Functions Solution

Find $(h j) (x)$ and $(\frac{j}{h}) (x)$ when $h (x) = x^{2} + 2 x - 15$ and $j (x) = x - 3 .$ Determine the domain for each function.

$(h j) (x) = h (x) \cdot j (x) (h j) (x) = (x^{2} + 2 x - 15) (x - 3) (h j) (x) = x^{3} - 3 x^{2} + 2 x^{2} - 6 x - 15 x + 45 (h j) (x) = x^{3} - x^{2} - 21 x + 45$

$(\frac{j}{h}) (x) = \frac{j (x)}{h (x)}, h (x) \neq 0 (\frac{j}{h}) (x) = \frac{x - 3}{x^{2} + 2 x - 15} = \frac{x - 3}{(x - 3) (x + 5)}, x \neq 3, - 5 (\frac{j}{h}) (x) = \frac{1}{x + 5}, x \neq - 5, 3$

${Domain}_{h} \{x | x \in ℝ\} {Domain}_{j} \{x | x \in ℝ\} {Domain}_{h j} \{x | x \in ℝ\} {Domain}_{\frac{j}{h}} \{x | x \in ℝ, x \neq - 5, 3\}$

Note

Q: How can you determine the excluded values for the denominator?

A: Factor the denominator and solve for x.

Q: Do you need to list restrictions that simplify out of the problem? Explain.

A: Yes, because those values would still make the denominator zero in the original problem.

Q: What is the value of (hj)(0)?

A: 45