How do I find local maxima and minima of a function?

1 Answer
Mar 22, 2018

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