Part 1 of 4 Walch the video and then solve the problem given below. Click here to watch the video. For the given functions, find \( (f \circ g)(x) \) and \( (g \circ f)(x) \) and the domain of each. \( f(x)=\frac{2}{1-7 x}, g(x)=\frac{1}{x} \) \( (f \circ g)(x)=\square \)

We are given:   f(x) = 2⁄(1 – 7x)   g(x) = 1⁄x To compute (f ∘ g)(x), we substitute g(x) into f(x):   (f ∘ g)(x) = f(g(x)) = f(1⁄x)     = 2⁄[1 – 7(1⁄x)]     = 2⁄(1 – 7⁄x) To simplify, write the denominator as a single fraction:   1 – 7⁄x = (x⁄x – 7⁄x) = (x – 7)⁄x Now, substitute back:   f(g(x)) = 2⁄[(x – 7)/x] = 2 · (x/(x – 7)) = 2x⁄(x – 7) So, we have:   (f ∘ g)(x) = 2x⁄(x – 7) Next, we must consider the domain. The domain of a composition is determined by:  1. The domain of g(x)  2. g(x) must be in the domain of f For g(x) = 1⁄x, we must have:   x ≠ 0 For f(x) = 2⁄(1 – 7x), the denominator (1 – 7x) ≠ 0, so:   1 – 7x ≠ 0 ⟹ x ≠ 1⁄7 But in f(g(x)) we are plugging in g(x) for x, so we need to check when:   1 – 7(g(x)) = 1 – 7(1⁄x) ≠ 0 Solve for when it equals zero:   1 – 7⁄x = 0 ⟹ 7⁄x = 1 ⟹ x = 7 Thus, for the composition:  • x ≠ 0 (because g(x) is not defined there)  • x ≠ 7 (because f(1⁄x) would be undefined when 1 – 7(1⁄x) = 0) Therefore, the domain of (f ∘ g)(x) is all real numbers x such that:   x ∈ ℝ \ {0, 7} Final answer:   (f ∘ g)(x) = 2x⁄(x – 7), with domain x ≠ 0 and x ≠ 7.