$$ f(x)=\frac{1}{9} x^{2}, x \leq 0 $$ Determine the equation of $$ f^{-1} $$ with restriction in the form $$ y=\ldots \ldots $$ On the same system of axes, sketch the graphs $$ f $$ and $$ f^{-1} $$. Indicate clearly the intercepts with axes, as well as another point on the graph of each of $$ f $$ and Is $$ f^{-1} $$ a function? Give a reason for your answer.

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1. **Find the inverse function $$ f^{-1} $$:**  
   We start with  
   $$
   y=\frac{1}{9}x^2,\quad x\le0.
   $$  
   To find the inverse, interchange the roles of $$ x $$ and $$ y $$:  
   $$
   x=\frac{1}{9}y^2.
   $$  
   Solve for $$ y $$:
   - Multiply both sides by 9:
     $$
     9x=y^2.
     $$
   - Take the square root:
     $$
     y=\pm\sqrt{9x}= \pm 3\sqrt{x}.
     $$
  
   Since the original function has domain $$ x\le 0 $$ (so its output $$ f(x) $$ is nonnegative) and is one‑to‑one, its inverse must return values that are in the original domain. Thus, for the inverse we choose the branch that yields non‑positive values:
   $$
   y=-3\sqrt{x}.
   $$
   Therefore, the equation of the inverse function is  
   $$
   f^{-1}(x)=-3\sqrt{x},\quad x\ge0.
   $$

2. **Intercepts and additional points:**
   
   - **For $$ f(x)=\frac{1}{9} x^2 $$ (with $$ x\le0 $$):**  
     - **Intercept with the $$ y $$-axis:**  
       Substitute $$ x=0 $$:
       $$
       f(0)=\frac{1}{9}(0)^2=0.
       $$
     - **Intercept with the $$ x $$-axis:**  
       Set $$ y=0 $$:
       $$
       \frac{1}{9}x^2=0\quad\Rightarrow\quad x=0.
       $$
     - **Another point:**  
       Let $$ x=-9 $$ (note $$ -9\le0 $$):
       $$
       f(-9)=\frac{1}{9}(-9)^2=\frac{81}{9}=9.
       $$
       So, the point $$ (-9,9) $$ lies on the graph of $$ f $$.

- **For $$ f^{-1}(x)=-3\sqrt{x} $$ (with $$ x\ge0 $$):**  
     - **Intercept with the $$ y $$-axis:**  
       Substitute $$ x=0 $$:
       $$
       f^{-1}(0)=-3\sqrt{0}=0.
       $$
     - **Intercept with the $$ x $$-axis:**  
       Set $$ y=0 $$:
       $$
       -3\sqrt{x}=0\quad\Rightarrow\quad \sqrt{x}=0\quad\Rightarrow\quad x=0.
       $$
     - **Another point:**  
       Let $$ x=9 $$ (since $$ 9\ge0 $$):
       $$
       f^{-1}(9)=-3\sqrt{9}=-3\cdot3=-9.
       $$
       So, the point $$ (9,-9) $$ lies on the graph of $$ f^{-1} $$.

3. **Sketching the graphs on the same coordinate system:**

- The graph of $$ f(x)=\frac{1}{9}x^2 $$ is a parabola opening upward but only the branch with $$ x\le0 $$ is considered.  
   - The graph of $$ f^{-1}(x)=-3\sqrt{x} $$ is the reflection of the graph of $$ f $$ about the line $$ y=x $$. It is defined for $$ x\ge0 $$ and produces non-positive outputs.
   - Both graphs intersect at the origin $$ (0,0) $$ and are symmetric with respect to the line $$ y=x $$.  
   - The point $$ (-9,9) $$ on $$ f $$ reflects to the point $$ (9,-9) $$ on $$ f^{-1} $$.

4. **Determine if $$ f^{-1} $$ is a function:**  
   Yes, $$ f^{-1} $$ is a function. For each input $$ x\ge0 $$ there is exactly one output, given by  
   $$
   f^{-1}(x)=-3\sqrt{x}.
   $$
   This one-to-one correspondence confirms that $$ f^{-1} $$ satisfies the definition of a function.
The inverse function is $$ f^{-1}(x) = -3\sqrt{x} $$ for $$ x \geq 0 $$. Both $$ f $$ and $$ f^{-1} $$ are functions as they pass the vertical line test.

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