Critical points of a cubic function There is a function $x^3 - 6x^2 + 9x + 1$.
Its critical points are $1$ and $3$.
I am very confused, if these points are maximum and minimum points respectively or are both inflection points. Can someone please help me with these ?
These points can not be maximum and minimum points since function attains higher and lower values as compared to what the function attains at these $2$ points. Please correct me if I am wrong.
 A: Indeed, a degree three polynomial has never maximum or minimum as it tends to plus and minus infinity. Yet, what is usually asked for in such contexts are local maximum and minimum. 
A way to find out if you have those is to consider the second derivative at those points. If it is negative it is a local  maximum, if it is positive it is a local minimum and if it is zero it is an inflection point.   
A: Let $f:x\mapsto x^3-6x^2+9x+1$, one has: $$f'(x)=3x^2-12x+9.$$
$1$ and $3$ are indeed critical points of $f$. In order to know their nature, one may compute $f''$, one has: $$f''(x)=6x-12.$$
In particular, $f''(1)=-6<0$ and $f''(3)=6>0$. Therefore, $1$ is a local minimum of $f$ while $3$ is a local maximum of $f$.
A: Those two are local maximum and minimum. It means that if you "restrict" your function to a certain small interval, then these two points would be "global" max/mins.
These are anyway critical points as the derivative vanishes at these points. Remember that when you have a function and you find the critical points, these could give you local min/max or a saddle point or also a global max min. Hence don't stop at finding the critical points.
You need to check the boundaries of the function and also you should use either the second derivative test or the hessian matrix, or other techniques to understand what kind of critical points they are.
A: Critical points are either local maxima, local minima or inflection points. For polynomials of odd degree they cannot be global maxima or minima unless the domain is restricted.

A: If the leading coefficient of the cubed term is positive (negative) and there are two distinct zeroes of the first derivative, then the smaller of the two is a local maximum (minimum) since the curve heads towards positive (negative) infinity.  The larger of the two zeroes would then be the local minimum (maximum).  The second derivatives are not needed, but would be negative for the local maximum and positive for the local minimum.
A: Take the derivative of $f(x) = x^3 - 6x^2 + 9x + 1$, which is $f'(x) = 3(x^2 - 4x + 3)$. Equal it to zero and you will get $x_1 = 1$ and $x_2 = 3$, which are the critical points. Now to classify them take the second derivative and it should be:
$f''(x) = 3(2x - 4)$. Now plug in the numbers and you will get that $f"(1) = -6 <0$ and $f''(3) = 6 > 0$. So by second derivative test $f(1)$ is a local minimum, but $f(3)$ is a local maximum
A: This theorem will allow you to easily calculate the nature of any critical point.
Suppose that at the critical point X, $$f'(X)=f''(X)=...=f^{(n-1)}(X)=0,$$ but $f^{(n)}(X)\neq0$, and that $f^{(n)}(X)$ is continuous in some neighborhood of X, where $n\geq2$. If n is even, then f has a local maximum at X if $f^{(n)}(X)<0$ and a local minimum if $f^{(n)}(X)>0$. If n is odd, then f has a horizontal inflection point at X.
We can test this on the function $f(x)=x^3$, which has a critical point at $x=0$. The derivatives at this point are $f'(0)=0$, $f''(0)=0$, $f'''(0)=6$. In this case, $n=3$, which is odd, therefore according to the theorem we have a horizontal inflection point.
