Question

Let g(x) = x - 3 and h(x) = x2 + 6. What is (h o g)(1)?

Answer 1

$\bf g(x)=x-3\qquad h(x)=x^2+6 \\\\\\ (h\circ g)(1)\implies h[\ g(1)\ ]\impliedby \textit{now, let's find g(1) first} \\\\\\ g(1)=(1)-3\implies g(1)=\boxed{-2}\qquad thus \\\\\\ h[\ g(1)\ ]\implies h\left( \boxed{ -2}\right)\implies h(-2)=(-2)^2+6 \\\\\\ h(-2)=4+6\implies h(-2)=10\impliedby (h\circ g)(1)$