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Prove that $$ f(x)=2x^4-8x+2$$ has exactly two real roots

I am aware of how to prove the existence of the two roots 1.4933.. and 0.2509.. by using Intermediate-Value theorem. However, I am unsure what theorem to use in showing that there is only two. I was thinking of going around proving that there is only one critical point and that it is the global minimum, but i am stuck there.

Sharmin Aeheli
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  • what happens if there was a 3rd root? if there can be no 3rd root then there can not be anymore than 2 roots – jimjim Sep 15 '21 at 14:16
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    Does Descartes' Rule of Signs solve this? Or perhaps you are looking for a more elementary solution. – Isky Mathews Sep 15 '21 at 14:17
  • I am looking for a method involving Rolles theoreom perhaps – Sharmin Aeheli Sep 15 '21 at 14:19
  • Please edit your question in order to give some background for the question. This is the kind of question which could (potentially) be asked in a precalculus class (with a largely qualitative answer being expected), in a calculus class (where the intermediate value and mean value theorems are relevant), or in a class on complex analysis (where Rouche might be relevant). – Xander Henderson Sep 15 '21 at 19:16
  • I would also suggest that you look at the "Related" questions to the right. Just reading over the topics, several of them seem to be almost identical to your question, with only some variation in constants. – Xander Henderson Sep 15 '21 at 19:17

2 Answers2

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Since $f'(x)=8(x^3-1)$, you know that $f$ is strictly decreasing in $(-\infty,1]$ and strictly increasing in $[1,\infty)$. Therefore, it cannot have more than two roots.

José Carlos Santos
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We have that $f(1)=-4$, $f(0)=2$, $f(2)=18$ and since

$$f'(x)=8x^3-8=0 \iff x=1$$

or simply since $f''(x)>0$, that is $f$ is convex, we can proceed to complete the proof by IVT.

user
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