Konichiwa has done the right thing by pointing out that f'(x) = 0 is only a condition of inflection and not necessarily one for an extremum. In investigating for maxima and minima, the points where f'(x) vanishes are termed critical points. They have to be investigated further for extremum conditions.

 

 A very simple example is f(x) = x³.

 

 

Proof:

Let us say the function f(x) has a maximum at a point a and is also differentiable there.

 

Since the function is differentiable the limits

 LHL =_h0  f(a) - f(a-h)/h and

RHL= _h0  f(a+h) - f(a)/h

exist and are equal.

 

But if x=a is a maximum point, then f(a) - f(x) 0 for points in the neighbourhood of a, Thus f(a) - f(a-h)0 and f(a+h) - f(a)0. Hence LHL0 and RHL0

 

But the derivative exists at the point and hence LHL = RHL. This can happen only when LHL=RHL=0. Thus f'(x) = 0 if derivative exists at the point of maximum.

A similar argument shows that f'(x) = 0 at a minimum point too.

 

But now can you work backwards to see if f'(a) =0, x=a is an extremum point?

 

1. Change of sign of f'(x) as it moves from x<a to x>a. (In effect f'(x) =0 at x=a)

2. f"(a) is non-zero. If f"(a)=0, the question of extremum remains open.

3. Even if f'(a) = f"(a) =....=fⁿ(a) = 0, but fⁿ⁺¹(a)  0. If n+1 is even then it is an extremum point. If n+1 is odd, there is no extremum at x=a.