Plotting tan(x)=x to show roots I am currently trying to plot tan(x)=x in Matlab but with the y=x portion only crossing the tan(x) at the intersection points.
Here is what I am trying to recreate in Matlab shown in Desmos:

As you can see, I have defined the domain to be $x,\:\left\{n\pi <x<\left(n+\frac{1}{2}\right)\pi \right\}$
Now here is my plot in Matlab:

As you can see, my lines do not cross the tan(x) like it dose in demos, but I cant really understand why. Here is my Matlab code:
f=@(x) tan(x); % defining an anyloums fucntion for tan(x)

%using a for loop to specifically show the y=x a lines of length of the
%given domian
for n=0:1:10;
    x=(n*pi:(n+1/2)*pi);
    y=x;
    plot(x,y)
    
    hold on;
  
    fplot(f,[0,11*pi])
   
end

The reason that I am doing this is because of a question that I have been given which is shown below.

By plotting the functions or otherwise, show that the equation tan(
x
) =
x
has an
infinite number of solutions $x_n$ where , $x_n,\:\left\{n\pi \le x_n\:\le \left(n+\frac{1}{2}\right)\pi \right\}$

 A: You have an interesting programmation error :
It comes from the fact that you have written 
x=n*pi:(n+1/2)pi 
Let us consider for example the case $n=1$. What this instruction generates is 
$x=[3.1416 , 4.1416]$
The lower bound is correct, the upper bound isn't (it should be $3\pi/2=4.7124$) 
Why that ? Because you have omitted to give a step like $0.01$ in  x=n*pi:0.01:(n+1/2)pi.
Thus, what happens ? The default step is one ; thus, what Matlab does is 
Begin by the left bound : $3.1416$, OK, then add the step 
$3.1416+1=4.1416$ OK because this value is below limit $4.7124$, but,
$3.1416+1+1=5.1416$ is above limit $4.7124$, thus isn't considered.
We are left with the two values $3.1416$ and $4.1416$...

Here is a way to correct your program :
clear all;close all;hold on;
axis([0,4*pi,-2,4*pi])
f=@(x) tan(x); 
x=0:0.01:4*pi;
plot(x,f(x))
for n=0:1:3;
    x=(n*pi:0.01:(n+1/2)*pi);
    y=x;
    plot(x,y)
end

You can observe that :


*

*the range has been transformed into  x=n*pi:0.01:(n+1/2)*pi; as announced.

*plotting $f$ has been externalized from your loop.

Besides :
You could have used a comma instead of a colon (very valuable remark by @Lutzl) :
x=[n*pi$\color{red}{,}$ (n+0.5)*pi]; y=x; plot(x,y);
I think that you can achieve a better vizualization by reconsidering your issue as the intersection of curves with equations 
$$y=\frac{\tan(x)}{x} \ \ \ \ \text{and} \ \ \ \  y=1$$
(see figure below) with the advantage to have points of interest (intersection points) all situated at the same level, closer and closer to the asymptotes with equations $(n+\frac12)\frac{\pi}{2}$.

A: For $n \in \mathbb N$ and $x \in (n\pi, n\pi +\pi/2)$, $\tan x-x$ is continuous, strictly increasing with $\tan n\pi-n\pi< 0$ and $\lim\limits_{x \to (n+1/2)\pi^-} \tan x-x=\infty$. Hence $\tan x-x$ has a unique root on this interval.
This is enough to conclude.
A: The equivalent equation $x=\arctan x+n\pi$ as fixed point iteration maps $\Bbb R$ to $[(n-\tfrac12)\pi,(n+\tfrac12)\pi]$, so as an endomorphism, self-map of the interval there has to be a fixed point. The next step tells that for $n>0$ the fixed point has to be inside the interval
$$
x>n\pi+\arctan((n-\tfrac12)\pi)=(n+\tfrac12)\pi-\arctan\left(\frac1{(n-\tfrac12)\pi}\right)>(n+\tfrac12)\pi-\frac1{(n-\frac12)\pi}\\
x<n\pi+\arctan((n+\tfrac12)\pi)=(n+\tfrac12)\pi-\arctan\left(\frac1{(n+\tfrac12)\pi}\right)<(n+\tfrac12)\pi 
$$
as $\arctan(x)<x$ for $x>0$. So you could draw even smaller intervals.
