Let #f(x)# be a real function of a real variable defined in#(a,b)# and differentiable in the point #x_0 in (a,b)#
A necessary condition for #x_0# to be a local minimum or maximum is that:
#f'(x_0) = 0#
If #f(x)# is differentiable in the entire interval, or at least in an interval around #x_0#, we also have a sufficient condition:
#x_0# is a local minimum if #f'(x_0) = 0# and there is a number #delta# such that:
#x in (x_0-delta, x_0) => f'(x) <=0#
#x in (x_0,x_0+delta) => f'(x) >=0#
In this case in fact #f(x)#will be decreasing on the left of #x_0# and increasing on the right, so that #x_0# is a relative minimum.
Vice versa #x_0# is a local maximum if:
#x in (x_0-delta, x_0) => f'(x) >=0#
#x in (x_0,x_0+delta) => f'(x) <=0#
In both cases the necessary condition is that the derivative of #f(x)# changes sign around #x_0#.
If #f(x)# also has a second derivative in an interval around #x_0# this is equivalent to the conditions:
#f'(x_0) = 0" and " f''(x_0) > 0 => x_0# is a local minimum.
#f'(x_0) = 0" and " f''(x_0) < 0 => x_0# is a local maximum.
So, to find local maxima and minima the process is:
1) Find the solutions of the equation:
#f'(x) = 0#
also called critical points.
2) Solve the inequality:
#f'(x) <=0#
to see if the sign of #f'(x)# changes around the critical points, or, alternatively:
2') Calculate #f''(x)# and look at its value in the critical points.
Example:
Let #f(x) = 2x^3-3x^2-12x+1#
Calculate the derivative:
#f'(x) = 6x^2-6x-12#
Solve the equation:
#f'(x) = 0#
# 6x^2-6x-12 =0#
# x^2-x-2 =0#
#x = (1+-sqrt(1+8))/2#
so the critical points are:
#x_1=-1#
#x_2 =2#
As #f'(x)# is a second degree polynomial with positive leading coefficients we know that:
#f'(x) < 0# for #x in (-1,2)#
#f'(x) > 0# for #x in (-oo,-1) uu (2,+oo)#
and we can soon determine that #x_1# is a local maximum and #x_2# a local minimum.
Alternatively we can evaluate the second derivative:
#f''(x) = 12x-6#
and check that:
#f''(x_1) = -12-6 = -18 < 0 => x_1# is a maximum
#f''(x_2) = 24-6 = 18 > 0 => x_2# is a minimum
