Calculus Critical Point Confusion - 2nd Derivative Confusion

Question

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.

Answer 1

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

Answer 2

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.

Answer 3

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.