Q. What is the best approximation of the minimum value attained by $e^{-x}\sin(100x)\ \ \quad$ such that $ x\ge 0$
My trial:
$$f(x)=e^{-x}\sin(100x)$$
$$f'(x)=100e^{-x}\cos(100x)-e^{-x}\sin(100x)$$
$$f''(x)=-9999e^{-x}\sin(100x)-200e^{-x}\cos(100x)$$
for minima,
$$f'(x)=0\implies100e^{-x}\cos(100x)-e^{-x}\sin(100x)=0$$
$$\tan(100x)=100\quad \because e^{-x}\ne0 $$
$$100x=k\pi+\tan^{-1}(100)=k\pi+1.56$$
$$x=\ldots-0.0158, 0, 0.0156, 0.04702, \dots\quad (\text{all values in radian})$$
here k is any integer. From graph there exist a number of minima of function $e^{-x}\sin(100x)$ but i don't know which values of k or $x\ge0$ should be taken to find the best approximation of minimum value. Somebody please help me solve it.
My book suggests that answer must lie in the range $-1$ to $-0.94$
thanks