Verified answer

Set f'(x)= 0, and find the critical numbers:

0= (x-1)^2 (x+2)^3

This may look overly complicated, but if you think about it, there are only two zeros: 1 and -2.

So, x = -2 and x = 1 are critical numbers.

Using that information, you can make up three intervals: (- infinity, -2); (-2, 1); and (1, infinity)

Pick an easy number to plug in from each interval into f'(x). So, -3, 0, and 2 should be fine.

x = -3

f'(-3) = (-3 - 1)^2 (-3 + 2)^3

f'(-3) = (16) (-1)

f'(-3) = -16

So, the function is decreasing on the interval (-infinity, -2)

x = 0

f'(0) = (0 - 1)^2 (0+2)^3

f'(0) = (1) (8)

f'(0) = 8

So, the function is increasing on the interval (-2, 1)

x = 2

f'(2) = (2-1)^2 (2+2)^3

f'(2) = (1) (64)

f'(2) = 64

So, the function is also increasing on the interval (1, infinity)

Because the function is decreasing on the interval (-infinity, -2) and increasing on the interval (-2, 1), there must be an absolute minimum at x = - 2.

When f'(x) = 0, f(x) has either a minimum or a maximum. f'(x) = 0 at x = -2 and 1

Now we can use sign pattern testing to see where f is increasing/decreasing.

When x < -2, f'(x) = (x-1)²(x+2)³ is negative, which means f is decreasing when x < -2

When -2 < x < 1, (x-1)²(x+2)³ is positive, which means f is increasing when -2 < x < 1

When x > 1, (x-1)²(x+2)³ is positive, which means f is increasing when x > 1

Since f is decreasing from negative infinity to -2, and then increasing from -2 to infinity, at x = -2, f has a minimum. x = -2 is also the absolute minimum because it is the ONLY minimum.

Differentiate f'(x) to get f''(x)

Find values of x where f''(x) changes sign from +ve to -ve or vice-versa.

One of the values will be -2....i.e. ur minimum.

Other will be ur maximum.