I must go to the higher derivatives?

[LagrangeEuler]

Function $f(x)=x^4$ has minimum at $x=0$.
$f'(x)=4x^3$
$f'(0)=0$
$f''(x)=12x^2$
$f''(0)=0$
$f^{(3)}(x)=24x$
$f^{(3)}(0)=0$
$f^{(4)}(0)>0$
So what is the rule? I must go to the higher derivatives if $f'(0)=f''(0)=0$?

 

[Mark44]

Function $f(x)=x^4$ has minimum at $x=0$.
$f'(x)=4x^3$
$f'(0)=0$
$f''(x)=12x^2$
$f''(0)=0$
$f^{(3)}(x)=24x$
$f^{(3)}(0)=0$
$f^{(4)}(0)>0$
So what is the rule? I must go to the higher derivatives if $f'(0)=f''(0)=0$?

No. You have f'(0) = 0 and f' changes from negative to positive as x changes, going left to right through zero. That's enough for you to conclude that there is at least a local minimum at zero.

Also, since f''(x) = 12x², which is positive everywhere except at x = 0, the graph of f is concave upward, so again there is a minimum at x = 0.

 

[HallsofIvy]

As Mark44 said, you can use the "first derivative test". But what you are saying is true. You can determine whether f(x) has a max or min by checking the first non-zero derivative at that point.

To see that think about writing the Taylor's polynomial for f up to that first non- zero derivative:
$$f(a)+ f'(a)(x- a)+ \frac{f''(a)}{2}(x- a)^2+ \cdot\cdot\cdot+ \frac{f^{(n)}}{n!}(x- a)^n$$
$$= f(a)+ \frac{f^{n}(a)}{n!}(x- a)^n$$
If n is odd x= a is a saddle point. If n is even, it is either a maximum or a minimum. If $f^n(a)> 0$ f increases away from x= a so that is a minimum. If $f^n(a)< 0$, it decreases away from x= a so that is a maximum.