SL 2.2—Functions, notation domain, range and inverse as reflection

The composition $(f\circ g)(4)$ means $f(g(4))$. First evaluate the inner function: $g(4)=\ln (4+5)=\ln 9$.

Now substitute into $f$: $f(g(4))=f(\ln 9)=e^{m\ln 9}$. Using the rule $e^{\ln a}=a$, this simplifies to $e^{m\ln 9}=9^m$.

We need $9^m=6$. Taking $\ln$ of both sides: $m\ln 9=\ln 6$, so $m=\frac{\ln 6}{\ln 9}$.

The value $m=\frac{\ln 6}{\ln 9}$ is correct. Note this can also be written as ${\log }_{9}6$.

We need to find $m$ such that $(f\circ g)(4)=6$. The composition gives $9^m=6$, so $m=\frac{\ln 6}{\ln 9}$.

If $m=\ln 6$, then $9^{\ln 6}\ne 6$. This option ignores the base from $g(4)=\ln 9$ entirely, so it is incorrect.

This would come from computing $g(4)=\ln 4$ instead of $\ln 9$. Since $g(4)=\ln (4+5)=\ln 9$, this option uses the wrong argument, so it is incorrect.

This would arise from evaluating $g(4)=\ln 5$ (i.e., using $4+1$ instead of $4+5$). This is a misreading of the function definition, so it is incorrect.

$m=\frac{\ln 6}{\ln 9}$ correctly follows from $g(4)=\ln 9$ and solving $9^m=6$.

The domain starts at $x=0$. Compute $f(0)=\sqrt{0+9}-3=3-3=0$. So the minimum output is $0$.

As $x$ increases, $x^2+9$ grows without bound, so $\sqrt{x^2+9}\to \infty$. Therefore $f(x)\to \infty$ as $x\to \infty$.

Since $x\ge 0$ and $\sqrt{x^2+9}$ is strictly increasing for $x\ge 0$, the function $f$ takes all values from $0$ upward continuously.

The range is $[0,\infty )$ — the function starts at $0$ and increases without bound.

The range is the set of all possible output values. We need to find what values $f(x)=\sqrt{x^2+9}-3$ can take for $x\ge 0$.

For $f(x)=-3$, we'd need $\sqrt{x^2+9}=0$, which is impossible since $x^2+9\ge 9>0$. So negative values in the range are impossible.

This would mean the minimum output is $3$. But $f(0)=0$, which is less than $3$. So $[3,\infty )$ is too restrictive.

This option suggests the range is bounded above by $9$. But as $x\to \infty$, $f(x)\to \infty$, so the range is unbounded. This option is incorrect.

$[0,\infty )$ is correct: the minimum value is $f(0)=0$ and the function grows without bound.