Steve S. answered 02/03/14
Tutor
5
(3)
Tutoring in Precalculus, Trig, and Differential Calculus
f(x) = x^7/2 - x^5/2 = x^(6/2 + 1/2) - x^(4/2 + 1/2) = (|x|^3 - |x|^2) √(x)
The constraint on the domain of f(x) comes from the radical.
For f(x) to be real, x ≥ 0. Then the absolute value symbols can be discarded:
f(x) = (x^3 - x^2) √(x), x ≥ 0.
g(x) = x^3/2 = x^(2/2 + 1/2) = x √(x), x ≥ 0.
( f + g)(x) = f(x) + g(x)
= (x^3 - x^2 + x) √(x), x ≥ 0.