Question 1199851
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If x = -1, then the denominator x+1 is zero.
Dividing by zero is NOT allowed. We must remove x = -1 from the domain.
In other words, {{{x <> -1}}}


The stuff under the square root is known as the radicand.
The radicand cannot be negative assuming you are currently studying real-valued functions only (ie. you have yet to study complex numbers)


So we want {{{(x-6)/(x+1)}}} to be nonnegative.
Symbolically we write {{{(x-6)/(x+1) >= 0}}}



Draw out a number line with -1 and 6 marked on it.
Sub-divide the number line into 3 non-overlapping regions<ul><li>Region A consists of stuff to the left of -1</li><li>Region B consists of stuff between -1 and 6</li><li>Region C consists of stuff to the right of 6</li></ul>Pick a random value from region A.
Let's go with x = -2
{{{(x-6)/(x+1)=(-2-6)/(-2+1) = (-8)/(-1) = 8}}}
The actual value doesn't matter. All we care about is whether it's positive or negative.
We got a positive result when x = -2
Thus any x value inside region A results in a positive radicand.
Furthermore, it means anything inside region A is part of the domain.


Pick a random value from region B.
Let's go with x = 0
{{{(x-6)/(x+1)=(0-6)/(0+1) = (-6)/1 = -6}}}
Any x value inside region B results in a negative radicand.
Stuff in region B is NOT in the domain.


Pick a random value from region C.
Let's go with x = 7
{{{(x-6)/(x+1)=(7-6)/(7+1) = 1/8 = 0.125}}}
Again the actual value doesn't matter. We only care about the sign.
We got a positive result so any x value inside region C results in a positive radicand.
In short, anything in region C is also part of the domain.


Summary table
<table border = "1" cellpadding = "5"><tr><td>Region</td><td>Interval</td><td>Sign of (x-6)/(x+1)</td></tr><tr><td>A</td><td>-infinity < x < -1</td><td>positive</td></tr><tr><td>B</td><td>-1 < x < 6</td><td>negative</td></tr><tr><td>C</td><td>6 < x < infinity</td><td>positive</td></tr></table>
I recommend drawing out the sign chart number line.


We see that regions A and C make (x-6)/(x+1) positive
So the domain is the set of x values that either x < -1 or x > 6
But wait: we can plug in x = 6 since it causes h(x) to be zero.
The numerator can be zero.


So we should update the domain to be {{{x < -1}}} or {{{x >= 6}}}


The interval notation would be (-infinity, -1) U [6, infinity)
The square bracket is to include the endpoint "6"
Curved parenthesis exclude the endpoint
Always use a curved parenthesis with either infinity.


Graph:
{{{graph(400,400,-8,8,-5,5,-100,sqrt((x-6)/(x+1)))}}}
The left branch approaches the vertical asymptote x = -1. It never reaches this value.
The gap between branches is due to the work shown in region B (see above).
There is a closed filled in endpoint at (6,0) which is the only x intercept.
There are no y intercepts.


Graphing tools such as Desmos and GeoGebra are very useful as quick visual verification methods. 
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