The function f'(x) = (x^3 -2)/x^4. If f(1) = 3, find f(x).

Given the derivative f'(x) = (x^3 -2)
/x^4

We need to find f(x).

We
know that the integral of f'(x) = f(x).

==> f(x) =
intg (x^3-2)/x^4  dx

Let us simplify
f'(x).

==> f(x) = intg (x^3/x^4) - 2/x^4  
dx

             = intg (1/x) - 2x^-4  
dx

             = intg (1/x)dx - intg (2x^-4) 
dx

             = ln x - 2x^-3/-3 +
C

             = lnx + 2/3x^3  +
C

But we know that f(1) =
3

==> f(1) = ln1 + 2/3  + C =
3

==> f(1) = 0 + 2/3 + C =
3

==> C = 3 - 2/3 =
7/3

==> f(x) = lnx + 2/3x^3 +
7/3