Question

let (f(x)=7x - 7) and (g(x)=x^{2}-2x + 3). then ((fcirc g)(x)=) ((gcirc f)(x)=) question help: video

Explanation:

Step1: Recall composition formula

$(f\circ g)(x)=f(g(x))$.

Step2: Substitute $g(x)$ into $f(x)$

Given $f(x) = 7x - 7$ and $g(x)=x^{2}-2x + 3$, then $f(g(x))=7(x^{2}-2x + 3)-7$.
Expand: $7x^{2}-14x+21 - 7=7x^{2}-14x + 14$.

Step3: Recall composition formula for $(g\circ f)(x)$

$(g\circ f)(x)=g(f(x))$.

Step4: Substitute $f(x)$ into $g(x)$

Since $f(x)=7x - 7$ and $g(x)=x^{2}-2x + 3$, then $g(f(x))=(7x - 7)^{2}-2(7x - 7)+3$.
Expand $(7x - 7)^{2}=49x^{2}-98x + 49$, $-2(7x - 7)=-14x + 14$.
So $g(f(x))=49x^{2}-98x + 49-14x + 14+3=49x^{2}-112x+66$.

Answer:

$(f\circ g)(x)=7x^{2}-14x + 14$
$(g\circ f)(x)=49x^{2}-112x + 66$