Examine the following function for extreme values:
f(x) = (x – 3)5 (x + 1)6
Here f (x) = (x – 3)5 (x + 1)6
Differentiating (I) w.r.t. x, we get
f ' (x) = 5 (x – 3)4 (x + 1)6 + (x – 3)5 (x + 1 )5
∴ f ' (x) = (x – 3)4 (x + 1)5 [5 (x + 1) + 6 (x – 3)]
= (x – 3)4 (x + 1)5 (11 x – 13)
Now, 
(a) When x = 3
If x < 3 (slightly), f ' (x) = (+) (+) (+) = + ve
If x > 3 (slightly), f ' (x) = (+) (+) (+) = + ve
Hence f ' (x) does not change sign as x passes through 3.
∴ x = 3 is neither a point of maxima, nor a point of minima. 3 is a point of inflexion.
(b) When x = – 1
If x < – 1 (slightly), f ' (x) = (+) (–) (–) = + ve
If x > – 1 (slightly), f ' (x) = (+) (+) (–) = – ve
∴ f ' (x) changes from positive to negative as a passes through – 1
Hence x = – 1 is a point of local maxima and maximum value of the function at x = – 1 is f (– 1) = 0.
(c) When 
