Why is a point, b, an extremum of a function if f'(b)=0?

1 Answer
Nov 7, 2015

A point at which the derivative is 0 is not always the location of an extremum.

Explanation:

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

has f'(x) = 3(x-1)^2 = 3x^2-6x+3,

so that f'(1)=0.

But f(1) is not an extremum.

It is also NOT true that every extremum occurs where f'(x)=0
For example, both f(x) = absx and g(x)=root3(x^2) have minima at x=0, where their derivatives do not exist.

It IS true that if f(c) is a local extremum, then either f'(c)=0 or f'(c) does not exist.