You should remember that the use of derivatives may help you in finding the maximum or minimum values a function can take, hence, you need to determine `f'(x)`  and you need to solve for x `f'(x) = 0`  such that:

`f(x) = (1/2)^(x^2-8x+15) => ln f(x) = (x^2-8x+15)*ln(1/2)`

Differentiating both sides yields:

`1/(f(x))*f'(x) = (2x-8)*ln(1/2)`

`f'(x) = (2x-8)*ln(1/2)*((1/2)^(x^2-8x+15))`

You need to solve for x the equation `f'(x)=0`  such that:

`(2x-8)*ln(1/2)*((1/2)^(x^2-8x+15)) = 0`

Since `((1/2)^(x^2-8x+15)) > 0`  for `x in R` , hence `2x-8 = 0`  such that:

`2x - 8 = 0 => 2x = 8 => x = 4`

You need to evaluate f(4) such that:

`f(4) = (1/2)^(4^2-32+15) => f(4) = (1/2)^(31-32) => f(4) = (1/2)^(-1) = 1/(1/2) => f(4) = 2` .

Hence, evaluating the maximum value the function can take, under the given conditions, yields `f(4) = 2` .