g(x) = (x/2)+(x^2)sin(1/x) when x=/=0 and g(0)=0. Prove that g is NOT increasing over any open interval containing 0.
This is a bit confusing to me, if x<0 then g(x)<0 and if x>0 then g(x)>0, so how is it not increasing. I am guessing there is some proof using the fact that a function f that is differentiable on (a,b) is increasing if and only if f'(x)>=0 for all x in (a,b), but I dont see where g'(x) is ever less than 0.
This is a bit confusing to me, if x<0 then g(x)<0 and if x>0 then g(x)>0, so how is it not increasing. I am guessing there is some proof using the fact that a function f that is differentiable on (a,b) is increasing if and only if f'(x)>=0 for all x in (a,b), but I dont see where g'(x) is ever less than 0.