Do we get some information from the graphs?

[mathmari]

Hey!

Suppose we have the function $f=\sqrt{x_1}x_2$, with $x_1, x_2>0$.

I want to say something about the intercection of a vertival plane through the graph of the above function.

To find the intersection I have done the following:

Setting $x_2=1$, we get $y=\sqrt{x_1}$ for the vertival intersection parallel to the $x_1$-$y$-plane at $x_2=1$.

Setting $x_2=2$, we get $y=2\sqrt{x_1}$ for the vertival intersection parallel to the $x_1$-$y$-plane at $x_2=2$.

Setting $x_2=4$, we get $y=4\sqrt{x_1}$ for the vertival intersection parallel to the $x_1$-$y$-plane at $x_2=4$. The computed vertical intersections are shown at the following graph:
[desmos="-10,10,-10,10"]\sqrt{x};2\sqrt{x};4\sqrt{x};[/desmos]Is this correct? (Wondering) We consider the scalar function $\psi (\lambda )=f(\lambda x_1, \lambda x_2), \lambda \geq 0$ at the point $(x_1, x_2)=(1,1)$.

Do we get the following graph for the scalar function? (Wondering)

[desmos="-10,10,-10,10"]x\sqrt{x\space}\left\{x\ge0\right\};[/desmos]
Do we get some information from the above graphs? (Wondering)

 

[jvicens]

Hello! It looks like you have correctly found the vertical intersections for the function $f=\sqrt{x_1}x_2$ by setting $x_2$ to different values and solving for $y$. These intersections can be visualized on a graph as shown in your post.

As for the scalar function $\psi(\lambda)$, it is correct that we can consider it at the point $(x_1,x_2)=(1,1)$ and graph it as $x\sqrt{x}$ for $x\geq 0$. This is because when we substitute in the values of $x_1$ and $x_2$ into the original function $f=\sqrt{x_1}x_2$, we get $f=x_1x_2=x_1\cdot 1=x_1$. Therefore, the scalar function becomes $\psi(\lambda)=\lambda x_1$ and when we substitute in $(x_1,x_2)=(1,1)$, we get $\psi(\lambda)=\lambda$ which is just a straight line with slope $1$ passing through the origin.

From these graphs, we can see that as we increase the value of $x_2$, the vertical intersections move further away from the origin, indicating that the function $f$ becomes steeper. Similarly, as we increase the value of $\lambda$, the scalar function $\psi$ becomes steeper, indicating that it is growing at a faster rate. I hope this helps!