Hello again Luandato,

The great thing about this question is that once you know the rule for functions, this will become a breeze for you!

A quick functions refresher before we tackle this question:
"y operations" affect the entire function and are on the outside of the function. E.g.

$$2f\left(x\right)\ =\ 2\cdot f\left(x\right)\ for\ the\ entire\ function$$
or
$$if\ g\left(x\right)\ =\ 2f\left(x\right),\ then\ every\ value\ of\ g\left(x\right)\ is\ always\ 2\ \times\ f\left(x\right)$$
These operations directly affect the y component of the function no matter what the x is. You're basically taking the function as a whole and manipulating it.

"x operations" affect the x portion (input) of the function. For example:
$$f\left(x\right)\ =\ 2x\ +1,\ then\ f\left(5x^2\right)\ =\ 2\ \times\left(5x^2\right)\ +\ 1\ =\ 10x^2+1$$
Another example:
$$f\left(x\right)\ =\ 2x\ +1,\ then\ f\left(g\left(x\right)\right)\ =\ 2\ \times\left(g\left(x\right)\right)\ +\ 1$$
So basically, you replace each x with the new "input."
In some cases, these are called function transformations.

Now, the fun part, solving our question:

$$f\left(k\right)\ =\ \frac{k^5}{16}$$
So, $$f\left(2k\right)\ =\ \frac{\left(2k\right)^5}{16}=\frac{2^5\ \cdot\ k^5}{16}=\frac{32\cdot k^5}{16}=32\cdot\frac{k^5}{16}=32\cdot f\left(k\right)\ ,\ for\ all\ k$$

As you can see, it was very important to take the entire (2k) to the fifth power, otherwise we would have incorrectly transformed the function. Always, always, always put your new "input"/x in parentheses and apply the appropriate operations.

That leaves us with OA E.