Question 443978
It doesn't sound like your teach knows what they're talking about.

There are multiple ways to find the domains of these functions:

One find the inverse of the function's range (which is the original's domain)
Two solve for where x is defined.

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For {{{sqrt(-1-2x)}}} it is defined strictly where {{{-1-2x >= 0}}}
So solve for x to get {{{-1 >= 2x}}} {{{-1/2>=x}} {{{x<= -1/2}}}
From -{{{infinity}}} to {{{-1/2}}} is the domain.
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You can find the inverse as well.  {{{y = sqrt(-1-2x)}}} {{{y^2 = -1-2x}}}
{{{y^2 + 1 = -2x}}} {{{-1*(y^2+1)/2 = x}}}
Switch the xs and ys to get
{{{y = expr(-1/2)(x^2+1)}}}
Since the range of {{{(x^2+1)}}} is {{{x>=1}}} then if you multiplied all values by -1/2, you'd get {{{x<=-1/2}}} which is the same as what we had.
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I believe the argument you teacher may have tried to make is:

{{{sqrt(-1-2x) = sqrt(-1*(1+2x)) = sqrt(-1) * sqrt(1+2x)}}}
Since {{{sqrt(-1)}}} is not allowed, then you can't find the domain. But this is not true. {{{sqrt(-1*(1+2x))}}} is defined where {{{(1+2x)<= 0}}}, because a negative times a negative gives us a positive (which we can take the square root of). So {{{(1+2x) <= 0}}}  {{{ 1 < -2x}}} {{{-1/2 >= x}}} {{{x <= -1/2}}}... again gives us the same result.
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Thus you can find the domain. 
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Does this make sense? I hope this helped.