Name \( \begin{array}{l}\text { Composite Functions Practice } \\ f(x)=-2 x^{2}, g(x)=\frac{1}{2} x+4, h(x)=3 x-5, j(x)=x^{2}+1 \\ \text { Use the equations above to find each composite function. } \\ \)\begin{tabular}{l|l|l|}\hline \text { 1. }\( j(g(x))= \) & \text { 2. }\( g(f(x))= \) & \text { 3. }\( f(h(x))= \)\end{tabular}\end{array}.

Sure! Let’s dive into the world of composite functions using the given equations. 1. For \( j(g(x)) \): First, find \( g(x) = \frac{1}{2}x + 4 \). Now, plug this into \( j(x) = x^2 + 1 \): \[ j(g(x)) = j\left(\frac{1}{2}x + 4\right) = \left(\frac{1}{2}x + 4\right)^2 + 1 \] Expanding gives: \[ = \frac{1}{4}x^2 + 4x + 16 + 1 = \frac{1}{4}x^2 + 4x + 17 \] 2. For \( g(f(x)) \): First, find \( f(x) = -2x^2 \). Now, plug this into \( g(x) = \frac{1}{2}x + 4 \): \[ g(f(x)) = g(-2x^2) = \frac{1}{2}(-2x^2) + 4 \] Simplifying gives: \[ = -x^2 + 4 \] 3. For \( f(h(x)) \): First, find \( h(x) = 3x - 5 \). Now, plug this into \( f(x) = -2x^2 \): \[ f(h(x)) = f(3x - 5) = -2(3x - 5)^2 \] Expanding gives: \[ = -2(9x^2 - 30x + 25) = -18x^2 + 60x - 50 \] So, we have: 1. \( j(g(x)) = \frac{1}{4}x^2 + 4x + 17 \) 2. \( g(f(x)) = -x^2 + 4 \) 3. \( f(h(x)) = -18x^2 + 60x - 50 \)