AHL 2.7—Composite functions, finding inverse function incl domain restriction

$(f\circ g)(x)=f(g(x))$, so we apply $g$ first, then $f$. Here $g(x)=x^2-1$ and $f(x)=3x+2$.

$g(x)=x^2-1$. This becomes the input to $f$.

$f(g(x))=f(x^2-1)=3(x^2-1)+2=3x^2-3+2=3x^2-1$.

$(f\circ g)(x)=3x^2-1$.

We need $(f\circ g)(x)=f(g(x))=3(x^2-1)+2$. Expanding gives $3x^2-3+2=3x^2-1$.

This would result from $f(g(x))=3(x^2)+2$, which ignores the $-1$ inside $g(x)$. Incorrect.

$3x^2-1$ is obtained by correctly substituting $g(x)=x^2-1$ into $f$ and simplifying: $3(x^2-1)+2=3x^2-1$.

$(p\circ q)(18)=p(q(18))$. Apply $q$ first, then $p$.

$q(18)=2(18)-8=36-8=28$. Wait — let me recheck: $2(18)-8=36-8=28$. Hmm, but the correct answer is $4$, so let me recompute. Actually $q(18)=2(18)-8=28$, and $p(28)=\sqrt{28}$. Let me re-examine: if $q(x)=2x-8$ and $x=18$, then $q(18)=28$. The answer $4$ comes from $q(x)=x-8$ evaluated at $x=24$, or from $q(18)=2(18)-8=28$... Actually let's check $q(12)=2(12)-8=16$, $p(16)=4$. So the intended input is $x=12$, but the question says $18$. Let me restate: with $q(x)=2x-8$, $q(18)=28$, $p(28)=\sqrt{28}$. The correct answer should be $\sqrt{28}$.

$p(28)=\sqrt{28}=2\sqrt{7}$. The correct answer is $\sqrt{28}$.

$(p\circ q)(18)=\sqrt{28}$.

Compute $(p\circ q)(18)=p(q(18))=\sqrt{q(18)}=\sqrt{2(18)-8}=\sqrt{28}$.

$\sqrt{10}$ would require $q(18)=10$, i.e. $2(18)-8=10$, but $28\ne 10$. Incorrect.

$4$ would require $q(18)=16$, i.e. $2(18)-8=16$, but $28\ne 16$. Incorrect.

This would result from incorrectly applying $q$ after $p$: $q(p(18))=2\sqrt{18}-8$, which is not the same composition. Incorrect.

$\sqrt{28}$ is the correct result: $q(18)=2(18)-8=28$, then $p(28)=\sqrt{28}$.