Question

find $g(f(x))$.
$f(x) = x^2 - 1$
$g(x) = \frac{1}{x - 1}$
$g(f(x)) = \frac{?}{x^2 - \square}$

Explanation:

Step1: Substitute f(x) into g(x)

To find \( g(f(x)) \), we substitute \( f(x) = x^2 - 1 \) into \( g(x) \). So we replace \( x \) in \( g(x) = \frac{1}{x - 1} \) with \( f(x) \), which gives \( g(f(x))=\frac{1}{f(x)-1} \).

Step2: Substitute f(x) expression

Now substitute \( f(x)=x^2 - 1 \) into the denominator: \( f(x)-1=(x^2 - 1)-1 \). Simplify the denominator: \( (x^2 - 1)-1=x^2-2 \)? Wait, no, wait. Wait, \( g(x)=\frac{1}{x - 1} \), so \( g(f(x))=\frac{1}{f(x)-1} \), and \( f(x)=x^2 - 1 \), so \( f(x)-1=(x^2 - 1)-1=x^2-2 \)? Wait, no, that's wrong. Wait, no, \( g(x)=\frac{1}{x - 1} \), so when we substitute \( f(x) \) into \( g(x) \), we have \( g(f(x))=\frac{1}{f(x)-1} \), and \( f(x)=x^2 - 1 \), so \( f(x)-1=(x^2 - 1)-1=x^2 - 2 \)? Wait, no, the numerator is 1, so the numerator is 1, and the denominator is \( f(x)-1=(x^2 - 1)-1=x^2 - 2 \)? Wait, no, wait the problem shows \( g(f(x))=\frac{?}{x^2 - \square} \). Wait, let's re - do it.

Wait, \( g(x)=\frac{1}{x - 1} \), so \( g(f(x))=\frac{1}{f(x)-1} \). Since \( f(x)=x^2 - 1 \), then \( f(x)-1=(x^2 - 1)-1=x^2-2 \)? No, wait, \( (x^2 - 1)-1=x^2-(1 + 1)=x^2 - 2 \). But the problem's form is \( \frac{?}{x^2-\square} \). So the numerator is 1, and the denominator is \( x^2 - 2 \)? Wait, no, the numerator is 1, so the numerator (the "?") is 1, and the denominator's constant term: \( x^2-2=x^2 - 2 \), so the square is 2? Wait, no, let's check again.

Wait, \( g(f(x))=\frac{1}{f(x)-1}=\frac{1}{(x^2 - 1)-1}=\frac{1}{x^2-2} \). So the numerator is 1, and the denominator is \( x^2 - 2 \), so the "?" is 1, and the square is 2.

Wait, let's verify:

\( g(f(x))=g(x^2 - 1)=\frac{1}{(x^2 - 1)-1}=\frac{1}{x^2-2} \). So the numerator is 1, and the denominator is \( x^2 - 2 \). So the "?" is 1, and the square is 2.

Answer:

The numerator (the "?") is 1, and the denominator's constant term (the "□") is 2. But the question asks for the "?", so the answer for the "?" is 1.