The function \( \mathbf{f} \) is such that \( f(x)=4 x-1 \) and the function \( \mathbf{g} \) is such that \( g(x)=k x^{2} \) where \( \mathbf{k} \) is a constant. Given that \( \mathbf{f g}(2)=12 \). Find \( \mathbf{k} \) (correct to 4 dp ). \( \} \mathbf{k}= \)

To solve for \( k \), we start by applying the definitions of the functions. First, we calculate \( f(g(2)) \): 1. First, find \( g(2) \): \[ g(2) = k \cdot 2^2 = 4k. \] 2. Now substitute \( g(2) \) into \( f(x) \): \[ f(g(2)) = f(4k) = 4(4k) - 1 = 16k - 1. \] We know that \( f(g(2)) = 12 \), gives us the equation: \[ 16k - 1 = 12. \] 3. Solve for \( k \): \[ 16k = 12 + 1 = 13, \] \[ k = \frac{13}{16} = 0.8125. \] So the value of \( k \) is \( 0.8125 \) (correct to 4 decimal places).