Answer
a. $-2x^2-x-1$
b. $2x^2-17x+41$
c. $-11$
d. $15$
Work Step by Step
Given $f(x)=4-x$ and $g(x)=2x^2+x+5$, we have:
a. $(f\circ g)(x)=4-(2x^2+x+5)=-2x^2-x-1$
b. $(g\circ f)(x)=2(4-x)^2+(4-x)+5=32-16x+2x^2+9-x=2x^2-17x+41$
c. $(f\circ g)(2)=-2(2)^2-(2)-1=-11$
d. $(g\circ f)(2)=2(2)^2-17(2)+41=15$
