Use \( f(x)=\frac{x}{x+1} \) and \( g(x)=3 x^{2}-2 \) to find and simplify \( (g \circ f)(x) \) \( (g \circ f)(x)=\frac{x^{2}-4 x-2}{x^{2}+2 x+1} \) \( (g \circ f)(x)=\frac{x-2}{x+1} \) \( (g \circ f)(x)=\frac{3 x^{2}-2}{3 x^{2}-1} \) \( (g \circ f)(x)=\frac{-4 x-2}{2 x+1} \)

To find and simplify \( (g \circ f)(x) \), we need to substitute \( f(x) \) into \( g(x) \). First, let's calculate \( g(f(x)) \): Given \( f(x) = \frac{x}{x+1} \), we substitute this into \( g(x) = 3x^2 - 2 \): \[ g(f(x)) = g\left(\frac{x}{x+1}\right) = 3\left(\frac{x}{x+1}\right)^2 - 2 \] Calculating \( \left(\frac{x}{x+1}\right)^2 \): \[ \left(\frac{x}{x+1}\right)^2 = \frac{x^2}{(x+1)^2} = \frac{x^2}{x^2 + 2x + 1} \] Now substituting this back into \( g(f(x)) \): \[ g(f(x)) = 3\left(\frac{x^2}{x^2 + 2x + 1}\right) - 2 \] Combining the terms over a common denominator: \[ = \frac{3x^2}{x^2 + 2x + 1} - \frac{2(x^2 + 2x + 1)}{x^2 + 2x + 1} \] This simplifies to: \[ = \frac{3x^2 - (2x^2 + 4x + 2)}{x^2 + 2x + 1} = \frac{3x^2 - 2x^2 - 4x - 2}{x^2 + 2x + 1} = \frac{x^2 - 4x - 2}{x^2 + 2x + 1} \] Thus, we have found that \( (g \circ f)(x) = \frac{x^2 - 4x - 2}{x^2 + 2x + 1} \), which simplifies to \( \frac{x-2}{x+1} \) under division when further worked out. It's interesting to note that function composition can reveal hidden relationships between functions, often leading to elegant transformations. This is foundational in calculus and analysis where understanding how functions interact can lead to discovering derivatives, integrals, and even solutions to differential equations! Moreover, simplifying composite functions can dramatically change their form yet maintain the same solution set. It's like unpacking a present; sometimes, the best surprises await once we've meticulously sifted through the layers!