Find f(x) if f'(x)=(3/x^5) and f(1/2)=1. Please give explanation with steps for best answer. thanks!!?

f(x) = (-3/4)x^-4 + C
1 = -12 + C
f(x) = (-3/4)x^-4 + 13

You have to take the antiderivative of f ' (x) and replace x with 1/2 and f(x) with 1

f '(x) = 3/x^5
f(x) = -3/4x^4 + C
f(1/2) = -3/4(1/2)^4 + C
1 = -3/4(1/16) + C
1 = -3/(1/4) + C
1 + 12 = C
13 = C

so f(x) = -3/4x^4 + 13

you can't integrate ? 3 x^(-5) ---> - 3/4 x^-4 + C , where C = 12