Implicit function Theorem - I have an example which I think is wrong

Question

I am trying to understand the implicit function theorem. Please consider the following system of equations which I believe does not have a solution. \begin{align*} e^{-x} - y &= 0 \\ e^{3x} + 10 - y &= 0 \end{align*} Now, I find the Jacobian: $$ \begin {vmatrix} -e^{-x} & -1 \\ 3e^{3x} & -1 \end{vmatrix} $$ Let $D$ be the determinate of the Jacobian. $$ D = e^{-x} + 3e^{3x} \neq 0 $$ Therefore the implicit function theorem tells me that the original system of equations has a unique solution but it does not. What am I missing?

Answer 1

You have the system of equations of

$$e^{-x} - y = 0 \iff e^{-x} = y \tag{1}\label{eq1A}$$ $$e^{3x} + 10 - y = 0 \tag{2}\label{eq2A}$$

This does have a solution for $x$ and $y$. To see this, substitute \eqref{eq1A} into \eqref{eq2A} to get

$$\begin{equation}\begin{aligned} e^{3x} + 10 - e^{-x} & = 0 \\ e^{3x} - e^{-x} & = -10 \end{aligned}\end{equation}\tag{3}\label{eq3A}$$

Next, have

$$f(x) = e^{3x} - e^{-x} \tag{4}\label{eq4A}$$

Note $f(x)$ is a continuous function for all real $x$, with $f(0) = 0$ and $f(-3) \approx -20.1$, so by the intermediate value theorem, there's an $-3 \lt x \lt 0$ such that $f(x) = -10$. Using this in \eqref{eq1A}, you get a corresponding value of $y$. This shows there's a solution to the system of equations.

Further, note this solution is unique. This is because

$$f'(x) = 3e^{3x} + e^{-x} \gt 0 \tag{5}\label{eq5A}$$

showing $f(x)$ is a strictly increasing function, so the value determined for $x$ is unique, and thus the corresponding value for $y$ is also unique.