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As the title says,

Let $p_1 = (1,0), p_2 = (2,3)$

and $f(p_1) = f(p_2) = 0$

and those are the only points of $\mathbb{R}^2$ which make $f$ zero.

Prove that the gradient of $f$ is $(0,0)$ at $p_1, p_2$.

I want to say that if the gradient wasn't equal to zero, then there would be a direction $v$ for which there would be a directional derivative equal to zero. Then there must be a point $p_3 = p_1 + t v$ with $t \to 0$ where $f(p_3) = 0$

But I don't know how to write that in a formal way, or even if its valid!

Thanks in advance for your help!!

1 Answers 1

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If we knew that a local extremum occurred at $p_1$, we could conclude that the gradient vanishes here (assuming this to be a sufficiently nice function, with continuous first and second order partial derivatives). To establish that $f(p_1)=0$ is an extreme value, we can show that $f$ is either always non-negative or always non-positive.

To do this, suppose we have $p_3,p_4\in\mathbb{R}^2$ with $f(p_3)<0$ and $f(p_4)>0$. We can choose a curve $c\colon[0,1]\to\mathbb{R}^2$ so that $c(0)=p_3$, $c(1)=p_4$, and so that our curve does not pass through $p_1$ or $p_2$. Then $f\circ c\colon[0,1]\to\mathbb{R}$ is a continuous function so that $f(0)<0$ and $f(1)>0$. Can you make a conclusion about $f\circ c$ that violates our assumption about where $f$ vanishes?

Once we obtain our contradiction, we conclude that either $f\geq 0$ on $\mathbb{R}^2$ or $f\leq 0$ on $\mathbb{R}^2$. In either case, $p_1$ and $p_2$ give local extrema of $f$, so the gradient of $f$ is zero at these points.

Can you make any similar statements when we increase the number of points where $f$ vanishes? If $f$ is zero only at $p_1,\ldots,p_n$ for some $n\in\mathbb{N}$, can you still conclude that the gradient vanishes at these points? What if the zero set of $f$ is some more general subset $C\subset\mathbb{R}^2$ of the plane? Are there conditions on this subset that will allow us to conclude that the gradient vanishes at these points?

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    Hi Austin, thank you very much for you answer you were very clear! I would use the theorem of the intermediate value for g = f∘c, and get to an absurd! However what you say about a set of n points is very interesting, I get where you are going (like when the set is a curve) but don't have very much idea of "when" the set starts allowing that the gradient on those points isn't 0. Would you mind explaining? I find it very interesting but I think that exceeds my course notes!2017-02-13
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    Edit! I think I've got it! for example if C is a set like x^2 + y^2 = 1, then there wouldn't be a curve from for example (0,0) to (2,2) without crossing a point of C (now how to write formally this...)2017-02-13
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    @Michael You have exactly the right idea. Your circle splits $\mathbb{R}^2$ into two *path components*, meaning that we can't give a path from a point in one component to a point in the other without crossing the circle. As long as the complement of $C$ in $\mathbb{R}^2$ is path-connected, we can use the above argument to conclude that the gradient vanishes on $C$, but the argument fails if $\mathbb{R}^2-C$ isn't path-connected. The gradient might vanish on $C$, but it also might not.2017-02-13
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    Hi @Austin how are you? just wanted to tell you I got a 9 in my test and a similar question appeared! so double thanks to you!! for your amazing answer!2017-02-24
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    @Michael Nice job on your test!2017-02-25