Once you write $f$ in terms of $y$, you will get an expression of the form $y + \dfrac{3}{y}$ with $y>0$. Ask: for which positive $y$ does this become smallest?

For $y>0$, the expression $y$ and $\dfrac{3}{y}$ balance each other. When are these two terms equal? How does that relate to getting the smallest possible sum?

After finding the value of $y$ that minimizes $y + \dfrac{3}{y}$, remember that $y = 2^x$. Solve $2^x = y$ for $x$ using logarithms base 2.

In Desmos, type f(x) = 2^x + 3/(2^x) to graph the function over a reasonable window, such as $x$ from about $-3$ to $3$.

On the graph of $f(x)$, look for the lowest point (the bottom of the curve). Tap or click on that point, or use the graphing calculator’s built-in "minimum" feature, and read off the $x$-coordinate; that $x$-value is where $f(x)$ attains its minimum.

Let $y = 2^x$. Because $2^x$ is always positive, $y>0$ for all real $x$.

Then

So the problem becomes: For what positive value of $y$ is $y + \frac{3}{y}$ minimized? After that, we will convert $y$ back to $x$.

For $y>0$, we can use the AM-GM inequality: for positive numbers $a$ and $b$,

with equality only when $a=b$.

Here, let $a = y$ and $b = \dfrac{3}{y}$. Then

Multiply both sides by $2$:

Equality (the minimum) happens when $a=b$, that is when

(since $y>0$). So $y$ must be $\sqrt{3}$ at the minimum.

We defined $y = 2^x$, and we found that the minimum occurs when $y = \sqrt{3}$.

So we solve

Write $\sqrt{3}$ as $3^{1/2}$ and use logs base 2:

Therefore, $f(x)$ attains its minimum value when $x = \dfrac{1}{2}\log_2 3$. This corresponds to choice A.