Functions — Question 8

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Question 8

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

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

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

  • (c) Evaluate (f∘g)(6)(f \circ g)(6).

Original worksheet page 1: question and worked solution for 1-1-008
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Question 8 - Solution

(a) Composite function and domain

The composite function is (f∘g)(x)=f(g(x))=f(x−2)=1x−2.(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≥0⇒x≥2x - 2 \ge 0 \Rightarrow x \ge 2.

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

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

(b) Endpoint analysis

At x=2x = 2, (f∘g)(2)=12−2=10,(f \circ g)(2) = \frac{1}{\sqrt{2 - 2}} = \frac{1}{0}, which is undefined. Therefore, x=2x = 2 is excluded from the domain.

(c) Evaluation

Evaluate the composite at x=6x = 6: (f∘g)(6)=16−2=14=12.(f \circ g)(6) = \frac{1}{\sqrt{6 - 2}} = \frac{1}{\sqrt{4}} = \frac{1}{2}.

Answer: 12\boxed{\frac{1}{2}}

Original worksheet page 2: question and worked solution for 1-1-008

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