Question

Find the points of local maxima or local minima, if any, of the functions, using the first derivative test. Also, find the local maximum or local minimum values, as the case may be: f(x) = (x – 1) (x + 2)2

Answer 1

Given, as f(x) = (x – 1) (x + 2)2

On differentiating with respect to x, we get,

f‘(x) = (x + 2)2 + 2(x – 1)(x + 2)

= (x + 2) (x + 2 + 2x – 2)

= (x + 2) (3x)

For all the maxima and minima,

f’(x) = 0

= (x + 2) (3x) = 0

On solving the above equation we get

= x =0, – 2

At x = – 2 f’(x) changes from negative to positive

Since, x = – 2 is a point of Maxima

At x =0 f‘(x) changes from negative to positive

Since, x = 0 is point of Minima.

Thus, local min value = f (0) = – 4

Local max value = f (– 2) = 0.