Calculus help! Derivatives and absolute value...?

Question

okay, suppose the first derivative of a function is an absolute function:
-I x I + 3
What is the function?

Answer 1

let f'(x) = -I x I + 3

we know that |x| = +x when x>0 and
|x| = -x when x<0

f'(x) = -x +3 when x>0
and
f'(x) = x+3 when x<0

to find the function, you integrate to get
f(x) = -(x^2)/2+3x + constant when x>0
and
f(x) = (x^2)/2 +3x +another constant when x<0

Good luck

Answer 2

IF x>0 y=-x^2/2 +3x+C
if x<=0 y= x^2/2+3x+C
C is an arbitrary constant

Answer 3

f'(x) = -x +3 when x>0
and
f'(x) = x+3 when x<0

to find the function, you integrate to get
f(x) = -(x^2)/2+3x + constant when x>0
and
f(x) = (x^2)/2 +3x +another constant when x<0

Answer 4

(-x|x|)/2 + 3x + C

i think everyone above is wrong.. because then the derivative wouldn't have abs value in it? idk though. good luck anyway =D