<p>What I say below in my answer is what I can say without  having read Acemoglu and Autor's article (unfortunately I haven’t it),  an answer based  on what you report in your question, of course.</p>
<p><strong>I think your graph is correct.</strong></p>
<blockquote>
<p><span class="math-container">$0 &lt; I_L &lt; I_H &lt; 1$</span>; this condition always holds.</p>
</blockquote>
<p>This condition  implies that you have to consider only the part of the graph where <span class="math-container">$ 0 &lt; I_L  &lt; 1$</span> and  <span class="math-container">$0 &lt; I_H &lt; 1$</span>, that is the part in the blue dotted square in the graph below, the <span class="math-container">$(0,1)\times (0,1)$</span> square:</p>
<p><a href="https://i.sstatic.net/19WkmTk3.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/19WkmTk3.png" alt="enter image description here" /></a></p>
<p>Moreover, I have added the 45° green line, as the condition quoted above says that it is always <span class="math-container">$I_H&gt;I_L$</span>, so that <strong>the graph should be comprised in the part of the plane below the 45° green line.</strong></p>
<p>This condition is respected in your graph, in the <span class="math-container">$(0,1)\times (0,1)$</span> square.</p>
<p>Instead, this condition is not respected in the graph of  Acemoglu and Autor's paper you posted.</p>
<p>As I said, I didn't read the paper of Acemoglu and Autor but, on the basis of what you refer, I suppose there is an inaccuracy in the graph.</p>
<p><strong>According to the condition  <span class="math-container">$I_L &lt; I_H $</span>, the values of the functions for <span class="math-container">$I_H=0$</span> can't be strictly positive.</strong></p>
<p>Indeed, in your graph you have a <span class="math-container">$0$</span> value for the blue  function and a negative value for the red function.
As I show also below, it seems to me that using functions like those in your graph, everything fits.
<span class="math-container">$$***$$</span></p>
<p><strong>How can Acemoglu and Autor's  graph be qualitatively obtained?</strong></p>
<p>We have, in addition, the assumption:</p>
<blockquote>
<p>The paper does not exactly define <span class="math-container">$\alpha$</span>. However, what it says is
that <span class="math-container">$\alpha_L (i) / \alpha_M (i)$</span> and <span class="math-container">$\alpha_M (i) / \alpha_H (i)$</span>
are continuously differentiable and strictly decreasing, with <span class="math-container">$i \in
 [0,1]$</span>.</p>
</blockquote>
<p>How can we obtain a qualitative graph of the two functions, without assuming a particular function for the <span class="math-container">$\alpha_j(i)$</span>?</p>
<p>The graph of the two functions can be qualitatively depicted evaluating <span class="math-container">$I_L$</span> when <span class="math-container">$L_H=0$</span>, and calculating the derivatives and evaluating their signs.</p>
<p>I re-write below the two  equations of Acemoglu and Autor in implicit form, to use them in the following:</p>
<p><span class="math-container">\begin{equation}
f_1(I_L, I_H)=\frac{A_M \alpha_M (I_H) M}{I_H - I_L} - \frac{A_H \alpha_H (I_H) H}{1- I_H}=0
\tag{1}\end{equation}</span></p>
<p><span class="math-container">\begin{equation}
f_2 (I_L, I_H)=\frac{A_L \alpha_L (I_L) L}{I_L} - \frac{A_M \alpha_M (I_L) M}{I_H - I_L}=0
\tag{2}\end{equation}</span></p>
<p>Moreover, it could be useful to write the equations in terms of <span class="math-container">$I_L$</span> and <span class="math-container">$I_H$</span> as you did in your question:</p>
<p><span class="math-container">\begin{equation}
I_L = I_H - \frac{A_M}{A_H} \frac{\alpha_M(I_H)}{\alpha_H(I_H)} \frac{M}{H}(1-I_H)
\tag {1'}\end{equation}</span></p>
<p><span class="math-container">\begin{equation}
I_H= I_L + \frac{A_M}{A_L} \frac{\alpha_L (I_L)}{\alpha_M (I_L)} \frac{M}{L} I_L
\tag {2'}\end{equation}</span></p>
<p>I suppose, from your question,  that the variables  are all positive, so we can see from <span class="math-container">$(1')$</span> that for <span class="math-container">$I_H=0$</span> <span class="math-container">$I_L$</span> assumes a negative value, as in your graph (the red fuction).</p>
<p>From <span class="math-container">$(2')$</span> it can be easily seen that the function starts from the origin (the blue function) (of course, then,  we must exclude the origin as by assumption <span class="math-container">$I_L\neq0$</span> and <span class="math-container">$I_H\neq 0$</span> and of course equation <span class="math-container">$(2')$</span> is equivalent to <span class="math-container">$(2)$</span> for <span class="math-container">$I_L\neq 0$</span> only, and for <span class="math-container">$I_L=0$</span> function <span class="math-container">$(2)$</span> is not defined).</p>
<p>Now, <strong>let's calculate the derivatives, and evaluate their signs.</strong></p>
<p>For  equation <span class="math-container">$(2)$</span> I use the <strong>implicit function theorem</strong>.  I re-write it as:</p>
<p><span class="math-container">\begin{equation}
f_2 (I_L, I_H)= A_L \frac {\alpha_L(I_L)}{a_M(I_L)}  L(I_H-I_L) - A_M M   I_L=0
\tag{2''}\end{equation}</span></p>
<p>From <span class="math-container">$(2'')$</span> we can calculate the derivative (<span class="math-container">$I_{L2}$</span> is the function implicitly defined by <span class="math-container">$(2)$</span> or <span class="math-container">$(2'')$</span>)</p>
<p><span class="math-container">$$\frac{dI_{L2}}{dI_H}=-  \frac { \frac {\partial f_2}{\partial I_H}} {\frac{\partial f_2}{\partial I_L}}= \frac {- A_L \frac {\alpha_L(I_L)}{a_M(I_L)}L}{-A_L \frac {\alpha_L(I_L)}{_M(I_L)}L +(I_H-I_L)A_L L\left (\frac{\alpha_L(I_L)}{\alpha_M(I_L)}\right)'-A_M M}&gt;0 $$</span></p>
<p>where <span class="math-container">$\left(\frac{\alpha_L(I_L)}{\alpha_M(I_L)}\right)'$</span> is the derivative of the ratio with respect to <span class="math-container">$I_L$</span>, which is negative by assumption.</p>
<p>Analogously, we can calculate the derivative for equation <span class="math-container">$(1)$</span> using the implicit function theorem, or simply differentiating <span class="math-container">$(1')$</span>, obtaining that its sign is positive.</p>
<p>In conclusion, we can draw a graph similar to your graph.</p>
<p><span class="math-container">$$***$$</span></p>
<p>The point left to understand is <strong>whether the two functions cross in the <span class="math-container">$(0,1)\times (0,1)$</span> square</strong>, that is an equilibrium exists (in your particular example, yes).</p>
<p><strong>We can resort to the intermediate value theorem,</strong> applied to the difference function <span class="math-container">$g(I_H)$</span> between the two functions defined by <span class="math-container">$(1)$</span> and <span class="math-container">$(2)$</span> above, on the interval <span class="math-container">$[0,1]$</span>. That is, define the function:</p>
<p><span class="math-container">$$g(I_H)= I_{L1}(I_H)- I_{L2}(I_H)$$</span></p>
<p>where  <span class="math-container">$I_{L1}$</span> and <span class="math-container">$I_{L2}$</span> are the functions implicitly defined by <span class="math-container">$(1)$</span> and <span class="math-container">$(2)$</span>, (and explicitated in <span class="math-container">$(2')$</span> and <span class="math-container">$(1')$</span>), that is respectively  the red and blue functions in the graph.</p>
<p>For <span class="math-container">$I_H=0$</span> we have already seen that <span class="math-container">$g(I_H) &lt;0$</span>.</p>
<p>For <span class="math-container">$I_H=1$</span> we can show  that <span class="math-container">$g(I_H)&gt;0$</span>.</p>
<p>Indeed, from <span class="math-container">$(1')$</span> it is easily seen that for <span class="math-container">$I_H=1$</span> the function <span class="math-container">$I_{L1}=1$</span>  (the red function),  as  also in your graph.</p>
<p>From equation <span class="math-container">$(2')$</span>, we can see that, for <span class="math-container">$I_H=1$</span>, <span class="math-container">$I_{L2}&lt;1$</span> (the blue function) (I omit the details otherwise this post becomes cumbersome).</p>
<p>Therefore,  <span class="math-container">$g(I_H)$</span> is positive for <span class="math-container">$I_H=1$</span> .</p>
<p>As <span class="math-container">$g(I_H)$</span> is a continuous function, as difference of two continuous functions, <strong>we can conclude, from the intermediate value theorem, that there must be a point in <span class="math-container">$(0,1)$</span> where <span class="math-container">$g(I_H)=0$</span>,</strong> that is where the two red and blue functions in the graph cross.</p>

CC BY-SA 4.0
2024-06-09T15:03:56.480Z
2024-06-10T12:00:40.337Z
2024-06-10T12:00:40.337Z
BakerStreet
<p>I'm going through the paper of Acemoglu and Autor (2011, Handbook of Labor Economics)</p>
<p>Consider the following system of two equations:</p>
<p><span class="math-container">\begin{equation}
\frac{A_M \alpha_M (I_H) M}{I_H - I_L} = \frac{A_H \alpha_H (I_H) H}{1- I_H}
\end{equation}</span></p>
<p><span class="math-container">\begin{equation}
\frac{A_L \alpha_L (I_L) L}{I_L} = \frac{A_M \alpha_M (I_L) M}{I_H - I_L}
\end{equation}</span></p>
<p>Everything is exogenous (basically, the exogenous variables are calibrated to some arbitrary values. In other words, take them as constants), except for the two endogenous variables of the system, <span class="math-container">$I_L$</span> and <span class="math-container">$I_H$</span>.</p>
<p>What you shall know now is that <span class="math-container">$0 &lt; I_L &lt; I_H &lt; 1$</span>; this condition always holds.</p>
<p>It would be simple to plot the related curves in the <span class="math-container">$I_H$</span> and <span class="math-container">$I_L$</span> space if we did not have <span class="math-container">$\alpha()$</span>, which is a function depending on the endogenous.</p>
<p>The paper does not exactly define <span class="math-container">$\alpha$</span>. However, what it says is that <span class="math-container">$\alpha_L (i) / \alpha_M (i)$</span> and <span class="math-container">$\alpha_M (i) / \alpha_H (i)$</span> are continuously differentiable and strictly decreasing, with <span class="math-container">$i \in [0,1]$</span>.</p>
<p>The paper says that from the two equations above, it is possible to get the following graph:</p>
<p><a href="https://i.sstatic.net/JpiVc5j2.png" rel="noreferrer"><img src="https://i.sstatic.net/JpiVc5j2.png" alt="enter image description here" /></a></p>
<p>Could you please demonstrate the mathematical steps needed to produce this graph?</p>
<p>That's my attempt. I expressed the first and second equation in terms of <span class="math-container">$I_L$</span> and <span class="math-container">$I_H$</span>, respectively, as follows</p>
<p><span class="math-container">\begin{equation}
I_L = I_H - \frac{A_M}{A_H} \frac{\alpha_M(I_H)}{\alpha_H(I_H)} \frac{M}{H}(1-I_H)
\end{equation}</span></p>
<p><span class="math-container">\begin{equation}
I_H= I_L + \frac{A_M}{A_L} \frac{\alpha_M (I_L)}{\alpha_L (I_L)} \frac{M}{L} I_L
\end{equation}</span></p>
<p>Then, for graphing this two functions, I assumed <span class="math-container">$\alpha_H \equiv e^{-i}$</span>, <span class="math-container">$\alpha_M \equiv e^{-1.3i}$</span> and <span class="math-container">$\alpha_L \equiv e^{-1.9i}$</span>, such that <span class="math-container">$\alpha_M/\alpha_H$</span> and <span class="math-container">$\alpha_L/\alpha_M$</span> are both strictly decreasing in <span class="math-container">$i$</span>.</p>
<p>For the sake of the graphical representation, I set <span class="math-container">$I_H = x$</span>, <span class="math-container">$I_L = y$</span>, and the following parametrization: <span class="math-container">$A_L= 1$</span>, <span class="math-container">$A_M=1.25$</span>, <span class="math-container">$A_H=1.6$</span>, <span class="math-container">$L=40$</span>, <span class="math-container">$M=60$</span>, <span class="math-container">$H=50$</span>, such that</p>
<p><span class="math-container">\begin{equation}
y= x - (1.25/1.6)(e^{-0.3x}) (60/50) (1-x)
\end{equation}</span></p>
<p><span class="math-container">\begin{equation}
x= y + 1.25 (e^{0.6y}) (60/40)y
\end{equation}</span></p>
<p>and when I graph them, I get a reasonable result since the two curve cross at <span class="math-container">$I_H=0.541$</span> and <span class="math-container">$I_L=0.176$</span>. However, as you can see below (the red curve refers to the first function), my graph intersects the x and y axes differently compared to the paper. What I am doing wrong?
<a href="https://i.sstatic.net/2fdrDIrM.png" rel="noreferrer"><img src="https://i.sstatic.net/2fdrDIrM.png" alt="enter image description here" /></a></p>

CC BY-SA 4.0
2024-06-08T19:49:28.013Z
What do you mean by 'an answer from a reputable source'? I can elaborate  an answer, but it is a mathematical answer, the 'source' can be just some mathematics theorem...
BakerStreet
CC BY-SA 4.0
2024-06-08T20:29:56.597Z
I did not pay much attention when I placed the bounty. Your answer will certainly be welcomed
Maximilian
CC BY-SA 4.0
2024-06-08T20:36:26.480Z
Ok, :-). Tomorrow I hope I have the time to review and post it.
BakerStreet
CC BY-SA 4.0
2024-06-09T03:17:45.480Z
I am also curious, especially about the values at the corner. In equation (1), if x=0, then y < 0 (unless $\alpha_M(IH) < 0$). Similarly, in equation (2), if y=0, x=0 too. However, in the diagram shown, both corner values of (1) and (2) attain strictly positive values. Therefore, I think there must be some tricks added to your $\alpha$ functions.
teddi
CC BY-SA 4.0
2024-06-09T23:57:49.663Z
@Maximilian Thank you for accepting my answer. Your question was excellent, congratulations.
BakerStreet
<microeconomics><macroeconomics><mathematical-economics>
Challenging question on mathematical economics
Maximilian
# Challenging question on mathematical economics

(Asked by: username_0 on 2024-06-05)

I'm going through the paper of Acemoglu and Autor (2011, Handbook of Labor Economics)

Consider the following system of two equations:

\begin{equation}
\frac{A\_M \alpha\_M (I\_H) M}{I\_H - I\_L} = \frac{A\_H \alpha\_H (I\_H) H}{1- I\_H}
\end{equation}

\begin{equation}
\frac{A\_L \alpha\_L (I\_L) L}{I\_L} = \frac{A\_M \alpha\_M (I\_L) M}{I\_H - I\_L}
\end{equation}

Everything is exogenous (basically, the exogenous variables are calibrated to some arbitrary values. In other words, take them as constants), except for the two endogenous variables of the system, $I\_L$ and $I\_H$.

What you shall know now is that $0 < I\_L < I\_H < 1$; this condition always holds.

It would be simple to plot the related curves in the $I\_H$ and $I\_L$ space if we did not have $\alpha()$, which is a function depending on the endogenous.

The paper does not exactly define $\alpha$. However, what it says is that $\alpha\_L (i) / \alpha\_M (i)$ and $\alpha\_M (i) / \alpha\_H (i)$ are continuously differentiable and strictly decreasing, with $i \in [0,1]$.

The paper says that from the two equations above, it is possible to get the following graph:

[![enter image description here](https://i.sstatic.net/JpiVc5j2.png)](https://i.sstatic.net/JpiVc5j2.png)

Could you please demonstrate the mathematical steps needed to produce this graph?

That's my attempt. I expressed the first and second equation in terms of $I\_L$ and $I\_H$, respectively, as follows

\begin{equation}
I\_L = I\_H - \frac{A\_M}{A\_H} \frac{\alpha\_M(I\_H)}{\alpha\_H(I\_H)} \frac{M}{H}(1-I\_H)
\end{equation}

\begin{equation}
I\_H= I\_L + \frac{A\_M}{A\_L} \frac{\alpha\_M (I\_L)}{\alpha\_L (I\_L)} \frac{M}{L} I\_L
\end{equation}

Then, for graphing this two functions, I assumed $\alpha\_H \equiv e^{-i}$, $\alpha\_M \equiv e^{-1.3i}$ and $\alpha\_L \equiv e^{-1.9i}$, such that $\alpha\_M/\alpha\_H$ and $\alpha\_L/\alpha\_M$ are both strictly decreasing in $i$.

For the sake of the graphical representation, I set $I\_H = x$, $I\_L = y$, and the following parametrization: $A\_L= 1$, $A\_M=1.25$, $A\_H=1.6$, $L=40$, $M=60$, $H=50$, such that

\begin{equation}
y= x - (1.25/1.6)(e^{-0.3x}) (60/50) (1-x)
\end{equation}

\begin{equation}
x= y + 1.25 (e^{0.6y}) (60/40)y
\end{equation}

and when I graph them, I get a reasonable result since the two curve cross at $I\_H=0.541$ and $I\_L=0.176$. However, as you can see below (the red curve refers to the first function), my graph intersects the x and y axes differently compared to the paper. What I am doing wrong?
[![enter image description here](https://i.sstatic.net/2fdrDIrM.png)](https://i.sstatic.net/2fdrDIrM.png)

**Tags:** microeconomics, macroeconomics, mathematical-economics
> **Comments:**
> **username_1**: What do you mean by 'an answer from a reputable source'? I can elaborate  an answer, but it is a mathematical answer, the 'source' can be just some mathematics theorem...  
> **username_0**: I did not pay much attention when I placed the bounty. Your answer will certainly be welcomed  
> **username_1**: Ok, :-). Tomorrow I hope I have the time to review and post it.  
> **username_2**: I am also curious, especially about the values at the corner. In equation (1), if x=0, then y < 0 (unless $\alpha_M(IH) < 0$). Similarly, in equation (2), if y=0, x=0 too. However, in the diagram shown, both corner values of (1) and (2) attain strictly positive values. Therefore, I think there must be some tricks added to your $\alpha$ functions.  
> **username_1**: @Maximilian Thank you for accepting my answer. Your question was excellent, congratulations.  

---

## 1 answers

### Answer by username_1 on 2024-06-09. Score: 5 | Accepted answer

What I say below in my answer is what I can say without having read Acemoglu and Autor's article (unfortunately I haven’t it), an answer based on what you report in your question, of course.

**I think your graph is correct.**

> $0 < I\_L < I\_H < 1$; this condition always holds.

This condition implies that you have to consider only the part of the graph where $ 0 < I\_L < 1$ and $0 < I\_H < 1$, that is the part in the blue dotted square in the graph below, the $(0,1)\times (0,1)$ square:

[![enter image description here](https://i.sstatic.net/19WkmTk3.png)](https://i.sstatic.net/19WkmTk3.png)

Moreover, I have added the 45° green line, as the condition quoted above says that it is always $I\_H>I\_L$, so that **the graph should be comprised in the part of the plane below the 45° green line.**

This condition is respected in your graph, in the $(0,1)\times (0,1)$ square.

Instead, this condition is not respected in the graph of Acemoglu and Autor's paper you posted.

As I said, I didn't read the paper of Acemoglu and Autor but, on the basis of what you refer, I suppose there is an inaccuracy in the graph.

**According to the condition $I\_L < I\_H $, the values of the functions for $I\_H=0$ can't be strictly positive.**

Indeed, in your graph you have a $0$ value for the blue function and a negative value for the red function.
As I show also below, it seems to me that using functions like those in your graph, everything fits.
$$\*\*\*$$

**How can Acemoglu and Autor's graph be qualitatively obtained?**

We have, in addition, the assumption:

> The paper does not exactly define $\alpha$. However, what it says is
> that $\alpha\_L (i) / \alpha\_M (i)$ and $\alpha\_M (i) / \alpha\_H (i)$
> are continuously differentiable and strictly decreasing, with $i \in
> [0,1]$.

How can we obtain a qualitative graph of the two functions, without assuming a particular function for the $\alpha\_j(i)$?

The graph of the two functions can be qualitatively depicted evaluating $I\_L$ when $L\_H=0$, and calculating the derivatives and evaluating their signs.

I re-write below the two equations of Acemoglu and Autor in implicit form, to use them in the following:

\begin{equation}
f\_1(I\_L, I\_H)=\frac{A\_M \alpha\_M (I\_H) M}{I\_H - I\_L} - \frac{A\_H \alpha\_H (I\_H) H}{1- I\_H}=0
\tag{1}\end{equation}

\begin{equation}
f\_2 (I\_L, I\_H)=\frac{A\_L \alpha\_L (I\_L) L}{I\_L} - \frac{A\_M \alpha\_M (I\_L) M}{I\_H - I\_L}=0
\tag{2}\end{equation}

Moreover, it could be useful to write the equations in terms of $I\_L$ and $I\_H$ as you did in your question:

\begin{equation}
I\_L = I\_H - \frac{A\_M}{A\_H} \frac{\alpha\_M(I\_H)}{\alpha\_H(I\_H)} \frac{M}{H}(1-I\_H)
\tag {1'}\end{equation}

\begin{equation}
I\_H= I\_L + \frac{A\_M}{A\_L} \frac{\alpha\_L (I\_L)}{\alpha\_M (I\_L)} \frac{M}{L} I\_L
\tag {2'}\end{equation}

I suppose, from your question, that the variables are all positive, so we can see from $(1')$ that for $I\_H=0$ $I\_L$ assumes a negative value, as in your graph (the red fuction).

From $(2')$ it can be easily seen that the function starts from the origin (the blue function) (of course, then, we must exclude the origin as by assumption $I\_L\neq0$ and $I\_H\neq 0$ and of course equation $(2')$ is equivalent to $(2)$ for $I\_L\neq 0$ only, and for $I\_L=0$ function $(2)$ is not defined).

Now, **let's calculate the derivatives, and evaluate their signs.**

For equation $(2)$ I use the **implicit function theorem**. I re-write it as:

\begin{equation}
f\_2 (I\_L, I\_H)= A\_L \frac {\alpha\_L(I\_L)}{a\_M(I\_L)} L(I\_H-I\_L) - A\_M M I\_L=0
\tag{2''}\end{equation}

From $(2'')$ we can calculate the derivative ($I\_{L2}$ is the function implicitly defined by $(2)$ or $(2'')$)

$$\frac{dI\_{L2}}{dI\_H}=- \frac { \frac {\partial f\_2}{\partial I\_H}} {\frac{\partial f\_2}{\partial I\_L}}= \frac {- A\_L \frac {\alpha\_L(I\_L)}{a\_M(I\_L)}L}{-A\_L \frac {\alpha\_L(I\_L)}{\_M(I\_L)}L +(I\_H-I\_L)A\_L L\left (\frac{\alpha\_L(I\_L)}{\alpha\_M(I\_L)}\right)'-A\_M M}>0 $$

where $\left(\frac{\alpha\_L(I\_L)}{\alpha\_M(I\_L)}\right)'$ is the derivative of the ratio with respect to $I\_L$, which is negative by assumption.

Analogously, we can calculate the derivative for equation $(1)$ using the implicit function theorem, or simply differentiating $(1')$, obtaining that its sign is positive.

In conclusion, we can draw a graph similar to your graph.

$$\*\*\*$$

The point left to understand is **whether the two functions cross in the $(0,1)\times (0,1)$ square**, that is an equilibrium exists (in your particular example, yes).

**We can resort to the intermediate value theorem,** applied to the difference function $g(I\_H)$ between the two functions defined by $(1)$ and $(2)$ above, on the interval $[0,1]$. That is, define the function:

$$g(I\_H)= I\_{L1}(I\_H)- I\_{L2}(I\_H)$$

where $I\_{L1}$ and $I\_{L2}$ are the functions implicitly defined by $(1)$ and $(2)$, (and explicitated in $(2')$ and $(1')$), that is respectively the red and blue functions in the graph.

For $I\_H=0$ we have already seen that $g(I\_H) <0$.

For $I\_H=1$ we can show that $g(I\_H)>0$.

Indeed, from $(1')$ it is easily seen that for $I\_H=1$ the function $I\_{L1}=1$ (the red function), as also in your graph.

From equation $(2')$, we can see that, for $I\_H=1$, $I\_{L2}<1$ (the blue function) (I omit the details otherwise this post becomes cumbersome).

Therefore, $g(I\_H)$ is positive for $I\_H=1$ .

As $g(I\_H)$ is a continuous function, as difference of two continuous functions, **we can conclude, from the intermediate value theorem, that there must be a point in $(0,1)$ where $g(I\_H)=0$,** that is where the two red and blue functions in the graph cross.


---

