Question 878371
<pre>
{{{sqrt(5-x)}}}{{{""<""}}}{{{x+1}}}

Get 0 on the right.

{{{sqrt(5-x)-x-1}}}{{{""<""}}}{{{"0"}}}

In order for {{{sqrt(5-x)}}} to be a real number we require

{{{5-x}}}{{{"">=""}}}{{{"0"}}}

{{{5}}}{{{"">=""}}}{{{x}}} which is the same as   

{{{x}}}{{{""<=""}}}{{{5}}}

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

Next we look for critical numbers of

{{{sqrt(5-x)-x-1}}}{{{""<""}}}{{{"0"}}}

By solving {{{sqrt(5-x)-x-1}}}{{{""=""}}}{{{"0"}}}
           
           {{{sqrt(5-x)}}}{{{""=""}}}{{{x+1}}}

           {{{(sqrt(5-x))^2}}}{{{""=""}}}{{{(x+1)^2}}}

           {{{5-x}}}{{{""=""}}}{{{x^2+2x+1}}}
  
           {{{"0"}}}{{{""=""}}}{{{x^2+3x-4}}}

           {{{"0"}}}{{{""=""}}}{{{(x+4)(x-1)}}}

            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 

{{{sqrt(5-x)-x-1}}}{{{""<""}}}{{{"0"}}}

{{{sqrt(5-(-5))-(-5)-1}}}{{{""<""}}}{{{"0"}}}

{{{sqrt(10)+5-1}}}{{{""<""}}}{{{"0"}}}

{{{sqrt(10)+4}}}{{{""<""}}}{{{"0"}}}

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 

{{{sqrt(5-x)-x-1}}}{{{""<""}}}{{{"0"}}}

{{{sqrt(5-(0))-(0)-1}}}{{{""<""}}}{{{"0"}}}

{{{sqrt(5)-1}}}{{{""<""}}}{{{"0"}}}

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 

{{{sqrt(5-x)-x-1}}}{{{""<""}}}{{{"0"}}}

{{{sqrt(5-(2))-(2)-1}}}{{{""<""}}}{{{"0"}}}

{{{sqrt(3)-3}}}{{{""<""}}}{{{"0"}}}

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 

{{{sqrt(5-x)-x-1}}}{{{""<""}}}{{{"0"}}}

{{{sqrt(5-(-4))-(-4)-1}}}{{{""<""}}}{{{"0"}}}

{{{sqrt(5+4)+4-1}}}{{{""<""}}}{{{"0"}}}
    
{{{sqrt(9)+3}}}{{{""<""}}}{{{"0"}}}

{{{3+3}}}{{{""<""}}}{{{"0"}}}

{{{6}}}{{{""<""}}}{{{"0"}}}

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 

{{{sqrt(5-x)-x-1}}}{{{""<""}}}{{{"0"}}}

{{{sqrt(5-(1))-(1)-1}}}{{{""<""}}}{{{"0"}}}

{{{sqrt(4)-2}}}{{{""<""}}}{{{"0"}}}
    
{{{2-2}}}{{{""<""}}}{{{"0"}}}

{{{"0"}}}{{{""<""}}}{{{"0"}}}

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 

{{{sqrt(5-x)-x-1}}}{{{""<""}}}{{{"0"}}}

{{{sqrt(5-(5))-(5)-1}}}{{{""<""}}}{{{"0"}}}

{{{sqrt(0)-6}}}{{{""<""}}}{{{"0"}}}
    
{{{-6}}}{{{""<""}}}{{{"0"}}}

That's true so 5 is a solution.

So we darken it:


-------------------------o===========&#9679;
-7 -6 -5 -4 -3 -2 -1  0  1  2  3  4  5

So the solution is 

1 < x &#8804; 5

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

Edwin</pre>