Functions — Question 8

Question 8

Let $$f(x) = \frac{1}{x}, \qquad g(x) = \sqrt{x - 2}.$$

• Find the composite function $(f \circ g)(x)$ and state its domain.

• Determine whether $(f \circ g)(x)$ is defined at $x = 2$. Justify your answer.

• Evaluate $(f \circ g)(6)$.

Question 8 - Solution

(a) Composite function and domain

The composite function is $$(f \circ g)(x) = f(g(x)) = f\!\left(\sqrt{x - 2}\right) = \frac{1}{\sqrt{x - 2}}.$$

To determine the domain:

• The square root requires $x - 2 \ge 0 \Rightarrow x \ge 2$.

• The denominator cannot be zero, so $\sqrt{x - 2} \neq 0 \Rightarrow x \neq 2$.

Thus, the domain is $$\boxed{(2, \infty)}.$$

(b) Endpoint analysis

At $x = 2$, $$(f \circ g)(2) = \frac{1}{\sqrt{2 - 2}} = \frac{1}{0},$$ which is undefined. Therefore, $x = 2$ is excluded from the domain.

(c) Evaluation

Evaluate the composite at $x = 6$: $$(f \circ g)(6) = \frac{1}{\sqrt{6 - 2}} = \frac{1}{\sqrt{4}} = \frac{1}{2}.$$

Answer: $$\boxed{\frac{1}{2}}$$