Given \( f(x)=5 x^{2}-4 \) and \( g(x)=6-\frac{1}{2} x^{2} \), find the following expressions. \( \begin{array}{llll}\text { (a) }(f \circ g)(4) & \text { (b) }(g \circ f)(2) & \text { (c) }(f \circ f)(1) & \text { (d) }(g \circ g)(0)\end{array} \) \( \begin{array}{ll}\text { (a) }(f \circ g)(4)=\square & \text { (Simplify your answer.) }\end{array} \)
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The Deep Dive
To solve the expressions given, let's break down each function and apply the necessary transformations. First, we need to find \( (f \circ g)(4) \): 1. Calculate \( g(4) \): \[ g(4) = 6 - \frac{1}{2}(4^2) = 6 - \frac{1}{2}(16) = 6 - 8 = -2. \] 2. Now substitute this into \( f(x) \): \[ f(-2) = 5(-2)^2 - 4 = 5(4) - 4 = 20 - 4 = 16. \] So, \( (f \circ g)(4) = 16 \). Next, we find \( (g \circ f)(2) \): 1. Calculate \( f(2) \): \[ f(2) = 5(2^2) - 4 = 5(4) - 4 = 20 - 4 = 16. \] 2. Now substitute this into \( g(x) \): \[ g(16) = 6 - \frac{1}{2}(16^2) = 6 - \frac{1}{2}(256) = 6 - 128 = -122. \] So, \( (g \circ f)(2) = -122 \). Next, we find \( (f \circ f)(1) \): 1. Calculate \( f(1) \): \[ f(1) = 5(1^2) - 4 = 5(1) - 4 = 5 - 4 = 1. \] 2. Now substitute this back into \( f(x) \): \[ f(1) = 1 \text{ (already found)}. \] So, \( (f \circ f)(1) = 1 \). Finally, we find \( (g \circ g)(0) \): 1. Calculate \( g(0) \): \[ g(0) = 6 - \frac{1}{2}(0^2) = 6 - 0 = 6. \] 2. Now substitute this into \( g(x) \): \[ g(6) = 6 - \frac{1}{2}(6^2) = 6 - \frac{1}{2}(36) = 6 - 18 = -12. \] So, \( (g \circ g)(0) = -12 \). To summarize: \[ \begin{array}{ll} \text{(a)} & (f \circ g)(4) = 16,\\ \text{(b)} & (g \circ f)(2) = -122,\\ \text{(c)} & (f \circ f)(1) = 1,\\ \text{(d)} & (g \circ g)(0) = -12. \end{array} \]
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