Necessary condition for locally invertible by inverse function theorem

If $$f:R²$$ to $$R²$$ such that f(x,y)=$$(x³+3xy²-15x-12y,x+y)$$. Let S={(x,y) such that f is locally invertible at (x,y)}. Then S is

Answer: my approach, I had used the inverse function theroem and find that at $$R²/{x-y=1, x-y=-1}$$ function is locally invertible. But how can I guarantee that at $$x-y=1$$ and $$x-y=-1$$,function has no neighborhood in which function is invertible. function itself is $$C¹$$ and Non vanishing of Jacobian matrix is sufficient conditon. I am not able to proceed.

Sorry, I am learning latex.

For instance, choose any point $$p$$ on the line $$x-y=1$$, which we can denote as $$(c+1,c)\in\mathbb{R}^2$$.

Regard the output of $$f$$ as $$(f_1,f_2)$$, where $$f_1(x,y)=x^3+3xy^2-15x-12y$$ and $$f_2(x,y)=x+y$$.

Our arbitrarily small neighbourhood of $$p$$, we can call $$U$$.

We are looking for points $$p_1=(x_1,y_1)$$ and $$p_2=(x_2,y_2)$$, distinct points in $$U$$, for which the output of $$f$$ is the same.

Let's try having at least $$f_2(x_1,y_1)=f_2(x_2,y_2)$$, since $$f_2$$ is easier than $$f_1$$. This is equivalent to $$x_1+y_1=x_2+y_2$$. Both points therefore belong to the same line $$L:x+y=k$$. To make things even simpler, I'll adjust $$k$$ so that $$p$$ is on that line, i.e. $$k=(c+1)+c=2c+1$$.

This allows us to substitute $$y=2c+1-x$$ for any arbitrary point on $$L$$. This in turn allows us to parametrise the points on $$L$$. The $$x$$-coordinate uniquely identifies each point on $$L$$, and the corresponding $$y$$-coordinate is $$2c+1-x$$.

Plug that straight into $$f_1$$. One will get $$f_1(x,2c+1-x)=\cdots$$ I can't bother writing it out.

But using Wolfram Alpha, you can check that $$\dfrac{d}{dx}f_1(x,2c+1-x)=12 (c + c^2 - 2 c x + (-1 + x) x)$$, which is equal to $$0$$ when $$x$$ is equal to $$c+1$$. Furthermore, $$\dfrac{d^2}{dx^2}f_1(x,2c+1-x)=-12(2c-2x+1)$$, which is equal to $$12$$ when $$x$$ is equal to $$c+1$$.

Therefore, $$f_1(x,2c+1-x)$$ actually attains a local maximum at $$x=c+1$$.

No matter how small $$U$$ is,

you can find $$x_1$$ slightly smaller than $$c+1$$, and $$x_2$$ slightly larger than $$c+1$$,

such that $$f_1(x_1,2c+1-x_1)=f_1(x_2,2c+1-x_2)$$.

Then we have our choice of $$(x_1,y_1)$$, where $$y_1=2c+1-x_1$$. Likewise for $$x_2$$.

• Thankyou so much.
– Tony
Mar 6 at 13:37
• Do you know necessary condition, such that if derivatives is zero at p then p will not have locally invertible?
– Tony
Mar 6 at 13:40
• I'm afraid I can't think of one. Mar 7 at 2:59
• I had asked a similiar question related locally open map using Jacobian. If you can see it in my profile, please give some hint.
– Tony
Mar 7 at 3:44