Find the inverse of the one-to-one function t(x) = 3*sqrt(x + 6), and state the domain of the inverse.

Step-by-step solution

1. 1

   Record the domain and range of the original function. The radicand must be non-negative, so $x+6\ge0$ gives domain $x\ge-6$; and since $\sqrt{\ }$ returns non-negative values, the range is $t(x)\ge0$. These two facts will become the range and domain of the inverse, respectively — which is why they are worth writing down first.

2. 2

   Set $y=t(x)$ and swap the variables. Writing $y=3\sqrt{x+6}$ and interchanging $x$ and $y$:

   $x=3\sqrt{y+6}.$

   Swapping is what encodes "reverse the input and output"; solving for $y$ afterwards produces the inverse rule.

3. 3

   Isolate the radical before squaring. Divide by $3$ first:

   $\frac{x}{3}=\sqrt{y+6}.$

   Squaring while the $3$ is still outside would give $x^{2}=9(y+6)$ — the same answer here, but the habit of isolating the radical first prevents errors when the coefficient is added rather than multiplied.

4. 4

   Square and solve for $y$. Squaring both sides:

   $\frac{x^{2}}{9}=y+6\;\Longrightarrow\;y=\frac{x^{2}}{9}-6.$

   So the inverse rule is $t^{-1}(x)=\dfrac{x^{2}}{9}-6$.

5. 5

   Attach the domain restriction — it is part of the answer. The parabola $\frac{x^{2}}{9}-6$ is not one-to-one on all of $\mathbb{R}$; only the branch matching the range of $t$ counts:

   $t^{-1}(x)=\frac{x^{2}}{9}-6,\qquad x\ge0.$

   Check both compositions: $t\!\left(t^{-1}(4.2)\right)=4.2$ ✓ and $t^{-1}\!\left(t(3.3)\right)=3.3$ ✓. Without the restriction, $t^{-1}(-3)=-5$ would wrongly claim $t(-5)=-3$, whereas in fact $t(-5)=3$.