Determining if $x=c$ is an extreme point of a function $f:\mathbb{R}\to\mathbb{R}$, given its Taylor polynomial of degree 2 at point $x=a$.

Question

Let $f:\mathbb{R}\to\mathbb{R}$ be a function such that $f\in C^3(\mathbb{R})$. Its quadratic Taylor polynomial at $x=1$ is $5-4(x-1)-(x-1)^2$. Prove/disprove: $x=0$ is not an extreme point of $f$.

Hey everyone. I'm trying to brush up on my calculus. I've come across this simple question and I'm not sure how to approach it. I've been looking into posts about Taylor polynomials and its applications, in regards to extreme points, but I haven't found anything helpful with this question.

Does the function $f$ necessarily share its extreme points with its Taylor polynomial of order $k$? I have some counter examples so no. I thought this was a classic disproof question, but I'm not sure how to construct such function.

I would be very happy to hear your thoughts since I'm lost. Thank you

Answer 1

Consider the following polynomial function:

$$f:\mathbb{R}\to \mathbb{R}:x \mapsto 5-4\left(x-1\right)-\left(x-1\right)^2+\frac{2}{3}\left(x-1\right)^3 \; .$$

This clearly has the required second order Taylor polynomial and it also possesses an extremum in $x=0$.