Get 0 on the right.



In order for  to be a real number we require



 which is the same as   



So 5 is a critical number, since no value of x
can be greater than 5

Next we look for critical numbers of



By solving 
           
           

           

           
  
           

           

            x+4=0;  x-1=0

              x=-4;   x=1

So there are two more critical numbers

We plot the three critical numbers on a number line:

----------o--------------o-----------o
-7 -6 -5 -4 -3 -2 -1  0  1  2  3  4  5

We test a value in each of the three regions

For the leftmost region, we test x=-5 in 









The left side is a positive number, so that is false.

So the left region is not part of the solution:

----------o--------------o-----------o
-7 -6 -5 -4 -3 -2 -1  0  1  2  3  4  5

For the middle region, we test x=0 in 







The left side is a positive number, so that is false.

So the middle region is also not part of the solution:

----------o--------------o-----------o
-7 -6 -5 -4 -3 -2 -1  0  1  2  3  4  5

For the righmost region, we test x=2 in 







The left side is a negative number, so that is true.

So the rightmost region is part of the solution,
so we shade that interval:

----------o--------------o===========o
-7 -6 -5 -4 -3 -2 -1  0  1  2  3  4  5

Finally we test the critical numbers themselves to
see if they are solutions;

We test critical number -4 in 






    






That's false so critical number -4 is not a solution.

So we erase the open circle at -4:


-------------------------o===========o
-7 -6 -5 -4 -3 -2 -1  0  1  2  3  4  5


We test critical number 1 in 






    




That's false so critical number 1 is not a solution.

-------------------------o===========o
-7 -6 -5 -4 -3 -2 -1  0  1  2  3  4  5

We test critical number 5 in 






    


That's true so 5 is a solution.

So we darken it:


-------------------------o===========●
-7 -6 -5 -4 -3 -2 -1  0  1  2  3  4  5

So the solution is 

1 < x ≤ 5

That can be written in interval notation as (1,5]

Edwin