Once you know the product $3^x \cdot 3^{1-x}$, think about how to relate the sum $3^x + 3^{1-x}$ to their product. Can you use the AM-GM inequality, which compares the arithmetic mean and geometric mean of two positive numbers?

For two positive numbers with a fixed product, their sum is smallest when the two numbers are equal. Set $3^x$ equal to $3^{1-x}$ to find the $x$-value where the minimum occurs, then plug that $x$ back into $g(x)$.

In Desmos, enter the function as y = 3^x + 3^(1 - x) and make sure the graph of the curve is visible in the main window.

Zoom in or out until you can clearly see where the curve reaches its lowest point. Tap/click on the curve near that lowest point and use the "Minimum" feature (or drag along the curve) to display the coordinates of the minimum point; note the y-value there.

Type each of the four answer choices into Desmos as separate expressions (exactly as written) to see their decimal values, and compare those decimals to the minimum y-value from Step 2. The choice whose decimal matches the minimum y-value is the correct answer.

The function is

Both terms $3^x$ and $3^{1-x}$ are positive for all real $x$.

Look at their product:

So the two terms $3^x$ and $3^{1-x}$ always multiply to $3$.

For any two positive numbers $a$ and $b$, the AM-GM inequality says

which is equivalent to

Here, let $a = 3^x$ and $b = 3^{1-x}$. Then

Equality in AM-GM (i.e., the smallest possible sum) happens only when $a = b$, so here when

We will use this to find the exact minimum value.

First solve $3^x = 3^{1-x}$. Because the bases are the same, set the exponents equal:

At this $x$, both terms are equal and the sum is minimized.

Now simplify the lower bound from Step 2 using the product from Step 1:

so

Check the value at $x = \tfrac{1}{2}$:

Because this matches the lower bound, the minimum value of $g(x)$ is $2\sqrt{3}$.