## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

$(f\circ g) (x)=11$ and $(g\circ f) (x)=2$
Let us consider $f(x)=2x^3-3x^2+4x-1$ and $g(x)=2$ Then we have: $(f\circ g) (x) =f[g(x)]=f [2]=(2)^3-(3)(2)^2+4(2)-1=16-12+8-1=11$ and $(g\circ f) (x) = g[f(x)]=g(2x^3-3x^2+4x-1)=2$ Therefore, $(f\circ g) (x)=11$ and $(g\circ f) (x)=2$