Answer:
[tex]S_{1 } (x,y) = (0.333, 0.815)[/tex] (Absolute minimum) and [tex]S_{2} (x,y) = (-1, 2)[/tex] (Absolute maximum)
Step-by-step explanation:
The critical points are determined with the help of the First Derivative Test:
[tex]f'(x) = 3\cdot x ^{2} +2\cdot x -1[/tex]
[tex]3\cdot x^{2} + 2\cdot x - 1 = 0[/tex]
The critical points are:
[tex]x_{1} \approx 0.333[/tex] and [tex]x_{2} \approx -1[/tex]
The Second Derivative Test offers a criterion to decide whether critical point is an absolute maximum and whether is an absolute minimum:
[tex]f''(x) = 6\cdot x +2[/tex]
[tex]f''(x_{1}) = 3.998[/tex] (Absolute minimum)
[tex]f''(x_{2}) = -4[/tex] (Absolute maximum)
The critical points are:
[tex]S_{1 } (x,y) = (0.333, 0.815)[/tex] (Absolute minimum) and [tex]S_{2} (x,y) = (-1, 2)[/tex] (Absolute maximum)
