Yes, the fact that f'(x) means there are no critical points- no max, min, or saddle point, just as your graph shows.
the problem is y=(1/x)+[1/(x+1)] I have to sketch the graph.
Listed below are my work.
let y=f(x)
f'(x)=[-1/x(^2)]-[1/((x+1)^2)]
I made f'(x)=0 and found that x can't be solved. Does this mean there are no critical points?
Asymptotes at x=0 and x=-1
Attached is the graph I drew.
Is this correct? What happens if I can't solve for x?
Thank you!