Step-by-step Solution:

First, simplify the given function \( f(x) \) by factoring the numerator and the denominator: \[ f(x) = \frac{3x^2+4x+1}{x^2+3x+2} \] Numerator: \( 3x^2+4x+1 = (3x+1)(x+1) \) Denominator: \( x^2+3x+2 = (x+2)(x+1) \) For \( x \ge 0 \), \( x+1 \neq 0 \), so we can cancel the common factor: \[ f(x) = \frac{3x+1}{x+2} \] We are asked to find the derivative of the inverse function at \( y = 2 \), denoted as \( (f^{-1})'(2) \). The formula for the derivative of an inverse function is: \[ (f^{-1})'(y) = \frac{1}{f'(x)} \] where \( y = f(x) \). First, find the \( x \) value corresponding to \( y = 2 \): \[ \frac{3x+1}{x+2} = 2 \] \[ 3x+1 = 2(x+2) \] \[ 3x+1 = 2x+4 \] \[ x = 3 \] Note that \( x = 3 \) is in the domain \( [0, \infty) \). Next, compute the derivative \( f'(x) \): \[ f'(x) = \frac{d}{dx} \left( \frac{3x+1}{x+2} \right) \] Using the quotient rule: \[ f'(x) = \frac{3(x+2) - (3x+1)(1)}{(x+2)^2} \] \[ f'(x) = \frac{3x+6 - 3x - 1}{(x+2)^2} = \frac{5}{(x+2)^2} \] Evaluate \( f'(x) \) at \( x = 3 \): \[ f'(3) = \frac{5}{(3+2)^2} = \frac{5}{5^2} = \frac{5}{25} = \frac{1}{5} \] Finally, apply the inverse derivative formula: \[ (f^{-1})'(2) = \frac{1}{f'(3)} = \frac{1}{1/5} = 5 \]