For $f(x)=x-3$ and $g(x)=5 x^{2}-3$, find the following functions. a. $(f \circ g)(x) ; b .(g \circ f)(x) ; c .(f \circ g)(0) ; d .(g \circ f)(0)$ a. $(f \circ g)(x)=5 x^{2}-6$ (Simplify your answer.) b. $(g \circ f)(x)=5 x^{2}-30 x+42$ (Simplify your answer.) c. $(f \circ g)(0)=\square($ Simplify your answer. $)$
Solution
Step 1 ：The given functions are \(f(x) = x - 3\) and \(g(x) = 5x^2 - 3\).
Step 2 ：The composition of two functions is denoted as \((f \circ g)(x)\) and \((g \circ f)(x)\), which means applying \(f\) to the result of applying \(g\) to \(x\), and vice versa.
Step 3 ：To find \((f \circ g)(x)\), we substitute \(g(x)\) into \(f(x)\), which gives us \((f \circ g)(x) = f(g(x)) = f(5x^2 - 3) = 5x^2 - 3 - 3 = 5x^2 - 6\).
Step 4 ：Similarly, to find \((g \circ f)(x)\), we substitute \(f(x)\) into \(g(x)\), which gives us \((g \circ f)(x) = g(f(x)) = g(x - 3) = 5(x - 3)^2 - 3 = 5x^2 - 30x + 42\).
Step 5 ：For part c, we substitute \(x = 0\) into the function \((f \circ g)(x)\), which gives us \((f \circ g)(0) = 5*0^2 - 6 = -6\).
Step 6 ：For part d, we substitute \(x = 0\) into the function \((g \circ f)(x)\), which gives us \((g \circ f)(0) = 5*0^2 - 30*0 + 42 = 42\).
Step 7 ：Final Answer: \(\boxed{(f \circ g)(x) = 5x^2 - 6}\), \(\boxed{(g \circ f)(x) = 5x^2 - 30x + 42}\), \(\boxed{(f \circ g)(0) = -6}\), \(\boxed{(g \circ f)(0) = 42}\)
