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

  • $\begingroup$ Definitely not. $\endgroup$ Jan 21, 2019 at 3:12
  • $\begingroup$ @WilliamElliot Can you please give me a clue on how to disprove this? thank you $\endgroup$
    – Noy
    Jan 21, 2019 at 6:49
  • $\begingroup$ A counterexample is sufficient. $\endgroup$ Jan 21, 2019 at 6:51
  • $\begingroup$ $x = 0$ is not even an extreme point of the Taylor polynomial! And even if it would be: your last sentence is not true -- just add something like $(x-1)^4$ to the Taylor polynomial to obtain a polynomial which does not posses an extreme point in $x = 0$. $\endgroup$
    – gerw
    Jan 21, 2019 at 6:56
  • 1
    $\begingroup$ @gerw I know, that would be a counter-example for the statement that says $f$ has got an extreme point at $x=0$, but I’m required to give a counterexample for the claim that $f$ hasn’t got an extreme point at $x=0$.. $\endgroup$
    – Noy
    Jan 21, 2019 at 9:26

1 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 $x=0$.


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