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 \( g(x) \). Start with \( f(x) = \frac{x}{x+1} \). Now let's calculate \( g(f(x)) \): \[ g(f(x)) = g\left(\frac{x}{x+1}\right) = 3\left(\frac{x}{x+1}\right)^{2} - 2 \] Calculating \( \left(\frac{x}{x+1}\right)^{2} \): \[ \left(\frac{x}{x+1}\right)^{2} = \frac{x^{2}}{(x+1)^{2}} = \frac{x^{2}}{x^{2}+2x+1} \] Substituting this back into \( g(f(x)) \): \[ g(f(x)) = 3 \cdot \frac{x^{2}}{x^{2}+2x+1} - 2 \] Now, write 2 as a fraction with the same denominator: \[ g(f(x)) = \frac{3x^{2}}{x^{2}+2x+1} - \frac{2(x^{2}+2x+1)}{x^{2}+2x+1} \] \[ = \frac{3x^{2} - (2x^{2} + 4x + 2)}{x^{2}+2x+1} \] \[ = \frac{3x^{2} - 2x^{2} - 4x - 2}{x^{2}+2x+1} \] \[ = \frac{x^{2} - 4x - 2}{x^{2}+2x+1} \] So, the simplified result is: \[ (g \circ f)(x) = \frac{x^{2}-4x-2}{x^{2}+2x+1} \] And here’s some fun information as a cherry on top! Communication through mathematics has been a historical marvel! From ancient Babylonians who have used sophisticated algorithms to the rise of algebra in the Islamic Golden Age, mathematics forms the bedrock of countless scientific discoveries that shape our world today. It's no wonder we still marvel at functions like \( f(x) \) and \( g(x) \) that continue this legacy! Mind the common slip-ups when working with composite functions! Often, people rush through substitution or forget to handle their algebra with care. Always double-check your calculations and be cautious with the signs, especially when you’re combining terms. Taking a moment to verify can save you a lot of headaches down the line. Happy calculating!