Question

Show that \(f\) and \(g\) are inverses of each other by verifying that \(f[g(x)]=x\) and \(g[f(x)]=x\). $$ f(x)=x^{2}+1(x \leq 0) ; \quad g(x)=-\sqrt{x-1} $$

Short answer

We verified that f[g(x)] = x and g[f(x)] = x for their respective domains. For f[g(x)], we found that the domain is x ≥ 1, and for g[f(x)], the domain is x ≤ 0. Since both expressions simplify to x, this shows that \(f(x) = x^2 + 1\) and \(g(x) = -\sqrt{x-1}\) are inverses of each other.

Step-by-step solution

Identify the functions f and g

: Here, we have the functions f and g defined as: \( f(x) = x^{2}+1 \quad (x \leq 0) \\ g(x) = -\sqrt{x-1} \) Keep in mind that f(x) is defined for x less than or equal to 0.

Find f[g(x)]

: Now, let's find f[g(x)]. Replace the input x in the function f with the function g(x). \( f[g(x)] = f[-\sqrt{x-1}] \) Since f(x) is defined for x less than or equal to 0, we have to make sure the input here, -√(x-1), is less than or equal to 0.

Ensure the domain of f[g(x)]

: Verify if g(x), i.e., -√(x-1), is less than or equal to 0. \( -\sqrt{x-1} \leq 0 \\ \sqrt{x-1} \geq 0 \\ x - 1 \geq 0 \\ x \geq 1 \) As we can see, g(x) is less than or equal to 0 when x ≥ 1. So, we can proceed with f[g(x)].

Compute f[g(x)]

: Plug in -√(x-1) into f(x), considering the condition x ≤ 0. \( f[g(x)] = [-\sqrt{x-1}]^2 + 1 \\ f[g(x)] = (x-1) + 1 \\ f[g(x)] = x \) So, we have verified that f[g(x)] = x.

Find g[f(x)]

: Now, let's find g[f(x)]. Replace the input x in the function g with the function f(x). \( g[f(x)] = g[x^2 + 1] \)

Compute g[f(x)]

: Plug in x^2 + 1 into g(x). \( g[f(x)] = -\sqrt{(x^2 + 1) - 1} \\ g[f(x)] = -\sqrt{x^2} \\ g[f(x)] = -|x| \) Since f(x) is defined for x ≤ 0, we know that -|x| = x. Therefore, g[f(x)] = x.

Verify both the results

: We have now found that f[g(x)] = x and g[f(x)] = x, which verifies that f and g are inverses of each other.