# How can I use inverse/implicit function theorem to find a function and its inverse?

Let $f:\mathbb{R}^3 \to \mathbb{R}$ be the function $f(x,y,z) = xy^2e^z + z$.

Show that there exists a function $g(x,y)$ defined around $\textbf{x} =(-1, 2)$ s.t. $g(\textbf{x}) = 0$ and $f(\textbf{x},g(\textbf{x})) = -4$. Find $g_x(\textbf{x})$ and $g_y(\textbf{x})$

I'm not sure where to start or how in order to show the existence of $g$, but this is the chapter about inverse and implicit function theorems. Any help would be more than welcome!

• In your definition of $f$ it has three arguments, $x,y,z$. In the use of it there are only two as $g$ seems to be a scalar. – Ross Millikan Apr 26 '16 at 14:11

Short Answer: The implicit function theorem is what is needed here. Try applying it to the function $F:\mathbb{R}^2 \times \mathbb{R} \rightarrow \mathbb{R}, (x,y,z) \mapsto f(x,y,z) + 4$ with the point $(x_0,y_0,z_0) = (-1,2,0)$ and see if you can conclude.

Long Answer: The implicit function theorem is exactly what is needed here to prove that such a function $g$ exists. The implicit function theorem works roughly speaking as follows:

If $F:\mathbb{R}^n \times \mathbb{R}^m \rightarrow \mathbb{R}^m$ is a continously differentiable function and $(x_0,y_0) \in \mathbb{R}^{n+m}$ a point such that $F(x_0,y_0) = 0$ and the "rightmost" $m\times m$-matrix of the $m\times (n+m)$-Jacobian of $F$ at $(x_0,y_0)$ (see edit below for more details) is invertible (i.e. the determinant of the $m\times m$-matrix should be non-zero), then there are:

• open neighborhoods $U \in \mathbb{R}^n$ and $V \in \mathbb{R}^m$ with $(x_0,y_0) \in U\times V$

• a unique continously differentiable function $g: U \rightarrow V$ satisfying $g(x_0) = y_0$

such that

• $F(x,y) = 0 \Leftrightarrow y = g(x)$ for every $(x,y) \in U\times V$

or, in other words,

• $F(x,g(x)) = 0$ for every $x \in U$.

Even more, $g$ is differentiable with the following formula (can be derived by using the multidimensional chain rule):

• $\frac{\partial g}{\partial x}(x) = -(\frac{\partial F}{\partial y}(x,g(x)))^{-1} \cdot \frac{\partial F}{\partial x}(x,g(x))$,

where with the partial derivatives the corresponding parts of the Jacobian is meant.

Try applying this to the function $F:\mathbb{R}^2 \times \mathbb{R} \rightarrow \mathbb{R}, (x,y,z) \mapsto f(x,y,z) + 4$ with the point $(x_0,y_0,z_0) = (-1,2,0)$ and see if you can conclude. If not, give a comment and I will edit my answer and provide additional information.

Remark: The implicit function theorem is used for checking if a given equation or a system of equations is unique solvable, at least locally. It does so by giving a function $g$ whose graph is the solution set of the equation on a local domain.

In general, the most difficult part in applying the theorem to a problem like yours is to correctly rewrite the given conditions an apropriate equation $F = 0$ with a function $F$, such that, assuming the theorem can be applied to it, solving the equation will meet the condition in the problem (i.e. in your case $f(x,g(x)) = -4$, where $x = (-1,2)$). After finding a possible equation you are almost done, since for applying the theorem you just have to calculate the Jacobian of $F$ and look if the determinant of the "right" partial matrix is nonzero. If this is the case, you get the existence of the function $g$.

Edit: What is meant with "the right partial matrix" is, according to the notation of the statement of the theorem above, the matrix that is formed by the partial derivatives of the $m$ variables $y_{n+1},\ldots,y_{n+m}$ in a point $(x,y) = (x_1,\ldots,x_n,y_{n+1},\ldots,y_{n+m}) \in \mathbb{R}^n \times \mathbb{R}^m = \mathbb{R}^{n+m}$ of $F = (F_1,\ldots, F_m)$, i.e. the matrix

\begin{pmatrix} \frac{\partial F_1}{\partial y_{n+1}}(x,y) & \cdots & \frac{\partial F_1}{\partial y_{n+m}}(x,y) \\ \vdots & & \vdots \\ \frac{\partial F_m}{\partial y_{n+1}}(x,y) & \cdots & \frac{\partial F_m}{\partial y_{n+m}}(x,y) \end{pmatrix}

which is "part" of the Jacobian of $F$. Sometimes, this is just called the partial derivative of $F$ according to $y = (y_{n+1},\ldots,y_{n+m})$, as seen in the formula of the derivative of $g$ above.

In your case of $F(x,y,z) = f(x,y,z) + 4$, the Jacobian is just the gradient of $F$, and checking if the determinant of the matrix above is nonzero reduces to the case of checking if the single number $\frac{\partial F}{\partial z}(-1,2,0)$ is nonzero.

It is true that you only have to consider this part of the Jacobian, i.e. you don't have to calculate the whole Jacobian in order to check if you can apply the theorem, though it will be necessary if you afterwards want to calculate the derivatives of the function $g$ with the formula above.

The maximum amount of variables that can be expressed in terms of a function $g$ equals the dimension of the range of $F$. So, as a rule to remember it is always handy to first write down the domain $\mathbb{R}^{n+m}$ of $F$ as a product $\mathbb{R}^n \times \mathbb{R}^m$ such that the dimension of the right factor coincides with the dimension of the range of $F$ in order to keep the overview of what variables you have to look at.

• Thanks alot for responding with such a great and detailed anwer. First of I am defining $F:(x, y, z) \mapsto xy^2e^z + z + 4$ which have continious partial derivatives. I am a little bit confused about "the right partial matrix", my the theorem says nothing about it in my book, but what it says instead is to verify that the partial of $F$ with respect to $y$ is non zero localy, which should be alot easier than to write out the whole jacobian, if i got the theorem right. Hope i made myself clear and that i understood what you stated here. Thanks again! – Thomas Apr 27 '16 at 5:54
• No problem, you're welcome. I'm sorry for the vague statement of "right partial matrix", I tried to state the theorem as intuitive as possible because the formal statement of it can be confusing (at least, it was for me when I first saw it). I edited my answer above, I hope that it will answer the questions in your comment. – user331406 Apr 27 '16 at 12:17