Calculus Critical Point Confusion - 2nd Derivative Confusion I am working with the function: 
$f(x) = 4x^5 - 3x^4 + 2x^3 - x^2 + 3$. 
One of the critical points $f'(x) = 0$ occurs when $x=0$. Checking the second derivative, we see that $f''(0) = -2$. This would lead one to the conclusion that the point $x=0$ is a local maximum. 
However, looking at the plot of the function, we see that $x=0$ looks more like a saddle: 

So, I am just wondering how we could conclude it is a saddle point if the second derivative is (incorrectly?) saying it is a local max. 
Thanks.
 A: The graph can lead to bad evaluation, indeed in this case fince $f'(0)=0$ and $f''(0)<0$ the point is of course a local maximum for the function.

A: Graphs can deceive. You have $f(0)=3$ and the derivative is
$$
f'(x)=20x^4-12x^3+6x^2-2x
$$
Nothing here hints to the function having an inflection point at $0$. To the contrary,
$$
f'(x)=2x(10x^3-6x^2+3x-2)
$$
is positive for $-\delta<x<0$ and negative for $0<x<\delta$ (some $\delta>0$), because the term in parenthesis is $-2$ for $x=0$.
Therefore $0$ is a point of local maximum, which is confirmed by $f''(x)=80x^3-36x^2+12x-2$ and $f''(0)=-2$.
In order that $c$ is an inflection point for a polynomial function $f$, one needs that $c$ is a root of multiplicity greater than one and odd of the polynomial $f(x)-f(c)$: just write the Taylor expansion at $c$ to realize it.
A: You have to zoom in more to understand the whole picture:
At first glance it might seem like a saddle point:

Let’s zoom in more

This is starting to look like more of max than a saddle so let’s zoom in a little bit more:

This is definitely a maximum according to our last picture.
Remember that even though it is a small interval and the function increases almost immediately after it, there are infinite points around $0$ such that they give a lower value for our function which makes $f(0)$ greater than its neighbours.
Source: Desmos Graphing Calculator
