# Inverse Function Theorem to solve functions

Prove that for $$(x, y) \in \Bbb R^2$$

$$\begin{cases} x+y+\sin(xy)=2c \\ \sin(x^2 + y) = c^2 \\ \end{cases}$$

can give a solution for all $$c ∈ \Bbb R$$ close enough to the origin, by the inverse function theorem.

I am familiar with the IFT, its conditions, and the result. However, I am having a hard time seeing connection between proving the problem and IFT, let alone solving it.

• For $c\in\mathbb R$, define $f_c : \mathbb R^2 \to \mathbb R^2$ by $f_c(x,y) = (x+y+\sin(xy)-2c,\sin(x^2+y)-c^2)$. Now verify the conditions of the implicit function theorem with this function. – azif00 Feb 12 at 4:12
• got it, can you plz further explain how this function proves the system of equations has a solution once I prove that it satisfies the conditions? – james black Feb 12 at 6:49
• i need a little in depth explanation as i cant make the connection between the thm and how the system of equations have a solution – james black Feb 12 at 6:50

Consider the function $$f(x,y,c):=\bigl(x+y+\sin(xy)-2c, \ \sin(x^2+y^2)-c^2\bigr)\ .$$ We obviously have $$f(0,0,0)=(0,0)$$. This means that when $$c=0$$ then the point $$(x_0,y_0):=(0,0)$$ solves your two equations. The implicit function theorem says that under a certain "technical assumption" we have $$f(x_c,y_c,c)=(0,0)$$ for all $$c$$ near $$0$$ and uniquely defined $$x_c=\phi(c)$$, $$\>y_c=\psi(c)$$ with $$\phi$$, $$\psi\in C^1$$ and $$\phi(0)=\psi(0)=0$$. Therefore your equations will have a single solution $$(x_c,y_c)$$ also for $$c$$ near $$0$$, whereby $$x_c$$, $$y_c$$ depend differentiably on $$c$$, and $$x_0=y_0=0$$.
The "technical assumption" is that $$f\in C^1$$ near $$(0,0,0)$$ (obviously satisfied), and that $$\det\left[\matrix{{\partial f_1\over\partial x}&{\partial f_1\over\partial y} \cr {\partial f_2\over\partial x}&{\partial f_2\over\partial y}\cr}\right]_{(0,0,0)}\ne0\ .$$ Therefore you have to compute this determinant and to check whether it is $$\ne0$$.