Raymond B. answered 04/12/24
Math, microeconomics or criminal justice
f(x) = 2x^2-2x, g(x)=-3x +8
a) f(g(x)) = f(-3x+8) = 2(-3x+8)^2 -2(-3x+8) = 2(9x^2-48x+64) +6x-16 =18x^2-96x+128 +6x-16
= 18x^2 -90x +112
b) g(f(x)) = g(2x^2-2x) = -3(2x^2-2x)+8 =-6x^2+6x+8
c) f(f(x)) = f(2x^2-2x) = 2(2x^2-2x)^2 -2(2x^2-2x) = 2(4x^4 -8x^3+4x^2) -4x^2 +4x
= 8x^4 -16x^3 +8x^2 -4x^2 +4x
= 8x^4 -16x^3+4x^2+4x
d) g(g(x)) = g(-3x+8) = -3(-3x+8) + 8 =9x-24+8 = 9x-16
the calculations get a little tedious, easy to make mistakes. One way to check the answer is try an easy integer for x, maybe x=0, then find f(0) =0 and f(f(0)) = f(0)=0, compare that with substituting 0 for x in f(f(x)) = 8x^4-16x^3+4x^2+4x and you get zero again.
try that with a) and it works. Not guaranteed correct, but much more likely
d) g(0)=8 g(8) =-24+8=-16. It works for d)
b) works too for x=0