$$ f(x)=3^{x} $$ and $$ g(x)=\left(\frac{1}{3}\right)^{2 x} $$

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To analyze the functions $$ f(x) = 3^x $$ and $$ g(x) = \left(\frac{1}{3}\right)^{2x} $$, we can rewrite $$ g(x) $$ in a more convenient form.

The function $$ g(x) $$ can be rewritten as follows:

$$
g(x) = \left(\frac{1}{3}\right)^{2x} = \left(3^{-1}\right)^{2x} = 3^{-2x}
$$

Now we have:

- $$ f(x) = 3^x $$
- $$ g(x) = 3^{-2x} $$

### Properties of the Functions

1. **Exponential Growth and Decay**:
   - $$ f(x) = 3^x $$ is an exponential growth function. As $$ x $$ increases, $$ f(x) $$ increases rapidly.
   - $$ g(x) = 3^{-2x} $$ is an exponential decay function. As $$ x $$ increases, $$ g(x) $$ decreases rapidly.

2. **Intercepts**:
   - For $$ f(x) $$, the y-intercept occurs at $$ f(0) = 3^0 = 1 $$.
   - For $$ g(x) $$, the y-intercept occurs at $$ g(0) = 3^{-2 \cdot 0} = 3^0 = 1 $$.

3. **Behavior as $$ x \to \infty $$ and $$ x \to -\infty $$**:
   - As $$ x \to \infty $$, $$ f(x) \to \infty $$ and $$ g(x) \to 0 $$.
   - As $$ x \to -\infty $$, $$ f(x) \to 0 $$ and $$ g(x) \to \infty $$.

### Intersection Points

To find the points where $$ f(x) = g(x) $$:

$$
3^x = 3^{-2x}
$$

Taking the logarithm base 3 of both sides:

$$
x = -2x
$$

Solving for $$ x $$:

$$
x + 2x = 0 \implies 3x = 0 \implies x = 0
$$

At $$ x = 0 $$:

$$
f(0) = 3^0 = 1 \quad \text{and} \quad g(0) = 3^{-2 \cdot 0} = 1
$$

Thus, the functions intersect at the point $$ (0, 1) $$.

### Summary

- $$ f(x) = 3^x $$ is an exponential growth function.
- $$ g(x) = 3^{-2x} $$ is an exponential decay function.
- Both functions intersect at the point $$ (0, 1) $$.
The functions $$ f(x) = 3^x $$ and $$ g(x) = 3^{-2x} $$ intersect at the point $$ (0, 1) $$.

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