f’(x) =

f’’(x) =

For maxima and minima,

f'(x) = 0

(x + 2)² = 4

x² + 4x = 0

x(x + 4) = 0

x = 0, – 4

Now f’’(0) = 1 > 0

x = 0 is point of minima

f’’(– 4) = – 1 < 0

x = – 4 is point of maxima

Local max value = f(– 4) = – 6

Local min value = f(0) = 2