This is what i'm getting.
f'(4) = 1.
Let f be the function defined for all x > -5 and having the following properties. Find an expression for f(x).
i) f ''(x) = 1/ (x+5)^1/3 for all x in the domain of f
ii) the line tangent to the graph of f at (4,2) has an angle of inclination of 45 degress.
I know
f '(x) =(3/2) (x+5)^2/3 + C
f (x) =(9/10) (x+5)^5/3 + (about -5.49)x + C
Well solving for C in f'(x) I got about -5.49
Solving for C to get f(x) I got about 70.254
Is this correct? If it is, is there an easier way to express those numbers in exact terms?