Given the functions $f(x) = x^2$ and $g(x) = 2x - 1$, we need to find (a) $g(f(x))$, which is the composite function $g \circ f$ and (b) $f(g(x))$, which is the composite function $f \circ g$.

1. Problem Description

Given the functions $f(x) = x^2$ and $g(x) = 2x - 1$, we need to find (a) $g(f(x))$, which is the composite function $g \circ f$ and (b) $f(g(x))$, which is the composite function $f \circ g$.

2. Solution Steps

(a) Find $g(f(x))$.

We have $f(x) = x^2$ and $g(x) = 2x - 1$. To find $g(f(x))$, we substitute $f(x)$ into $g(x)$ in place of $x$.

$g(f(x)) = g(x^2)$

$g(x^2) = 2(x^2) - 1 = 2x^2 - 1$

Therefore, $g(f(x)) = 2x^2 - 1$

(b) Find $f(g(x))$.

We have $f(x) = x^2$ and $g(x) = 2x - 1$. To find $f(g(x))$, we substitute $g(x)$ into $f(x)$ in place of $x$.

$f(g(x)) = f(2x - 1)$

$f(2x - 1) = (2x - 1)^2$

$f(2x - 1) = (2x - 1)(2x - 1)$

$f(2x - 1) = 4x^2 - 2x - 2x + 1$

$f(2x - 1) = 4x^2 - 4x + 1$

Therefore, $f(g(x)) = 4x^2 - 4x + 1$

3. Final Answer

(a) $g(f(x)) = 2x^2 - 1$

(b) $f(g(x)) = 4x^2 - 4x + 1$