Critical Points problem of Calculus

y = e^x + sin x
y' = e^x + cos x

For x≧0, we should note that e^x is always greater than cos x, implying y is monotonic , i.e. y has no critical point for x≧0

For x<0, 0 < e^x < 1, monotonic and -1 < cos x < 1, periodic. Thus y' cuts through x-axis occasionally. So all the solutions for x to the equation e^x + cos x = 0 are the critical points. They are all stationary points.

(In fact, e^x is so small that can be neglected and y' ≒ cos x. A close estimation to the critical points are x ≒ -pi/2, -3pi/2, -5pi/2 ......)

Hope the above information helps =)
By 小儒