For $f_{a,b}(x)=2^{x+a-b}-abx+4$, show that for $x_1,x_2$ close enough to $2,3$ there are $a,b$ such that $f_{a,b}(x_1)=f_{a,b}(x_2)=0$

Question

Let $f_{a,b}(x)=2^{x+a-b}-abx+4$. Show that for $x_1$ close enough to $2$ and $x_2$ close enough to $3$ there are $a,b$ such that $f_{a,b}(x_1)=f_{a,b}(x_2)=0$.

Hint: $f_{2,2}(2)=f_{2,2}(3)=0$.

It appears to have a lot to do with the Implicit Function Theorem.

My problems are:

1. In my Professor's, quite unorthodox notations (Is saying "Unorthodox" disrespectful towards him? I am not native.), the theorem is defined for the origin, in which the function is $0$, and not just any point in the Euclidean space.

2. While the Theorem provides information about the neighborhood of the point in for which the function is zero, it doesn't provide much information regarding a neighborhood containing two points\solutions to $f(x)=0$. Thus I am left with the duty to show that there are two neighborhoods of $((a,b)(2),2)$ and $((a,b)(3),3)$ in which the function is zero, and that these neighborhoods intersect.

I am completely overwhelmed, and would benefit from your perspective on the field.

I have been thinking a little and this is my attempt:

Using the more conventional form of The Implicit Function Theorem, $f:\Bbb{R}^2\times\Bbb{R}\to \Bbb{R}$, is continuous in all its variables. So is its gradient. Therefore it is $C_1$ (Will that do?). Whether or not it was correctly shown that $f$ is $C_1$, assuming it is, since $f(2,2,2)=0$, there exists a neighborhood $V_{(2,2)}\times V_2\in \Bbb{R}^3$ (Meaning, an intersection of neighborhoods of $(2,2)$ and $2$ in $\Bbb{R}^3$) of $(2,2,2)$ in which $f(x,y)=0\iff y=g(x),x\in\Bbb{R}^2$. Thus, for $x_1$ close enough to $2$, $x_1\in V_2$, and we get a neighborhood $U_{x_1}$ of $(2,2)$, satisfying $U_{x_1}\subset V_{2,2}$ such that on $U_{x_1}\times V_2$, $f(x,x_1)=0\iff x_1=g(x), x\in \Bbb{R}^2$. Doing the same for $x_2$ close to $3$, we get a neighborhood $U_{x_2}\times V_3$ and $x_2$ is represented as a function of $a$ and $b$ in $U_{x_2}$. Since $U_{x_1}\cap U_{x_2}\ne \emptyset$, there are $a,b$ as required. Am I close?

Answer 1

Strategy: Add more dimensions to the domain and range of $f$ and then apply implicit function theorem.

Define $F:\mathbb{R^4\to\mathbb{R}^2}$ as $$ F(x_1,x_2,a,b)=\begin{bmatrix}2^{x_1+(a-b)}-abx_1+4\\2^{x_2+(a-b)}-abx_2+4\end{bmatrix}=\begin{bmatrix}F _1(x_1,x_2,a,b)\\F_2(x_1,x_2,a,b)\end{bmatrix}=\begin{bmatrix}f _{a,b}(x_1)\\f_{a,b}(x_2)\end{bmatrix}\\F(2,3,2,2)=\begin{bmatrix}0\\0\end{bmatrix}\\ J=\begin{bmatrix}{\partial F_1\over\partial a}&\partial F_1\over\partial b\\\partial F_2\over\partial a&\partial F_2\over\partial b\end{bmatrix}\Big{|}_{(2,3,2,2)}=\begin{bmatrix}4\ln2-4&-4\ln2-4\\8\ln2-6&-8\ln2-6\end{bmatrix}\\ \det(J)=16\ln2\neq0 $$ According to implicit function theorem $\exists$ open sets $U,V\subset\mathbb R^2$ containing $(2,3)$ and $(2,2)$ respectively and a unique $\mathcal C^1$ function $g:U\to V$ such that $$ (x_1,x_2)\in U\implies F(x_1,x_2,g(x_1,x_2))=0 $$ So if you pick $(x_1,x_2)$ close enough to $(2,3)$, $g(x_1,x_2)$ is the required $(a,b)$.