Given \( f(x)=5 \sqrt{x} \) and \( g(x)=5 x \), evaluate the following expressions. \( \begin{array}{llll}\text { (a) }(f \circ g)(4) & \text { (b) }(g \circ f)(2) & \text { (c) }(f \circ f)(1) & \text { (d) }(g \circ g)(0)\end{array} \) \( \begin{array}{l}\text { (a) }(f \circ g)(4)=\square \\ \text { (Type an exact answer, using radicals as needed. Simplify your answer.) }\end{array} \)
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Extra Insights
To evaluate the expressions, we first need to understand how function composition works. Let's go through each of the expressions step by step. (a) \((f \circ g)(4) = f(g(4))\) First, we need to find \(g(4)\): \[ g(4) = 5 \times 4 = 20 \] Now, plug this result into \(f\): \[ f(20) = 5 \sqrt{20} = 5 \sqrt{4 \times 5} = 5 \times 2\sqrt{5} = 10\sqrt{5} \] So, \((f \circ g)(4) = 10\sqrt{5}\). (b) \((g \circ f)(2) = g(f(2))\) First, we find \(f(2)\): \[ f(2) = 5 \sqrt{2} \] Now plug this into \(g\): \[ g(5\sqrt{2}) = 5 \times (5 \sqrt{2}) = 25 \sqrt{2} \] So, \((g \circ f)(2) = 25\sqrt{2}\). (c) \((f \circ f)(1) = f(f(1))\) First, calculate \(f(1)\): \[ f(1) = 5 \sqrt{1} = 5 \] Now plug this result into \(f\): \[ f(5) = 5 \sqrt{5} \] So, \((f \circ f)(1) = 5\sqrt{5}\). (d) \((g \circ g)(0) = g(g(0))\) First, calculate \(g(0)\): \[ g(0) = 5 \times 0 = 0 \] Next, plug this result into \(g\): \[ g(0) = 5 \times 0 = 0 \] So, \((g \circ g)(0) = 0\). Summarizing the results: - (a) \((f \circ g)(4) = 10\sqrt{5}\) - (b) \((g \circ f)(2) = 25\sqrt{2}\) - (c) \((f \circ f)(1) = 5\sqrt{5}\) - (d) \((g \circ g)(0) = 0\)
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