How do you determine if #f(x)=x+absx# is an even or odd function?

2 Answers
Apr 5, 2016

Relate #f(-x)# to #f(x)#

Explanation:

#f(-x) = -x + abs(-x)#

#= -x + abs(x)#

Since #f(-x) != f(x)#, #f(x)# is not an even function.
Since #f(-x) != -f(x)#, #f(x)# is not an odd function.

Here is a graph of #y = f(x)#.
graph{x+abs(x) [-10, 10, -5, 5]}
If #f(x)# is an even function, the #y#-axis would be a line of symmetry.

If #f(x)# is an odd function, the graph would have rotational symmetry about the origin.

These are graphical methods to check whether a function is odd or even. However neither of the symmetries are present.

Hence, #f(x)# is neither an odd function nor an even function.

Apr 5, 2016

Neither.

Explanation:

#f(-x) = -x + |x|# is neither f(x) nor -f(x).
#|x|=-x, x<0# and #|x|=x, x>0#.
#f(x)=x-x=0, x<=0#..
#f(x)=2x, x>0#..
The graph for y = f(x) comprises the negative x-axis continued as the straight line y = 2x in the first quadrant.,