The function $$f\left( x \right) = \int\limits_{ - 1}^x {t\left( {{e^t} - 1} \right)\left( {t - 1} \right){{\left( {t - 2} \right)}^3}{{\left( {t - 3} \right)}^5}dt} $$         has a local minimum at $$x = ?$$

Solution :
$$\eqalign{ & \frac{{dy}}{{dx}} = f'\left( x \right) \cr & \Rightarrow x\left( {{e^x} - 1} \right)\left( {x - 1} \right){\left( {x - 2} \right)^3}{\left( {x - 3} \right)^5} = 0 \cr} $$
Critical points are $$0,\,1,\,2,\,3$$
Consider change of sign of $$\frac{{dy}}{{dx}}$$  at $$x = 3.$$
$$x < 3,\,\frac{{dy}}{{dx}} = $$   negative and $$x > 3,\,\frac{{dy}}{{dx}} = $$   positive
Change is from negative to positive, hence minimum at $$x = 3.$$  Again minimum and maximum occur alternately.
$$\therefore $$  2nd minimum is at $$x = 1.$$