Question 1208034
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You are correct.
The domain of the inverse is the range of the original function.


The domain and range swap places like this because x and y swap. After the swap the goal is to solve for y.
f(x) = (x^2+2)/(x+4)
y = (x^2+2)/(x+4)
x = (y^2+2)/(y+4) .... swapping x and y
x(y+4) = y^2+2
xy+4x = y^2+2
y^2-xy+2-4x = 0


To finish the process of solving for y, we use the quadratic formula.
a = 1
b = -x
c = 2-4x
{{{y = (-b+-sqrt(b^2-4ac))/(2a)}}}


{{{y = (-(-x)+-sqrt((-x)^2-4*1*(2-4x)))/(2*1)}}}


{{{y = (x+-sqrt(x^2+16x-8))/(2)}}}
The presence of "plus/minus" indicates the inverse isn't a function. 
Notice how the original function f(x) fails the horizontal line test.


We want the radicand x^2+16x-8 to be nonnegative so we avoid the square root of a negative number.
Apply the quadratic formula to x^2+16x-8 = 0 to find the exact solutions are {{{x = -8-6*sqrt(2)}}} and {{{x = -8+6*sqrt(2)}}} 


Graph out y = x^2+16x-8
The graph is a parabola that opens upward.
The portion between the roots is when y < 0, so the radicand x^2+16x-8 is only nonnegative when {{{x<=-8-6*sqrt(2)}}} and {{{x>=-8+6*sqrt(2)}}}


Therefore the range of the function f(x), as a set of inequalities, would be {{{y<=-8-6*sqrt(2)}}} or {{{y>=-8+6*sqrt(2)}}}


The range in interval notation would be 
{{{
drawing(400,100,-5,5,-5,5,
arc(-3,-1,3,5,140,220),
locate(-4,0,matrix(1,3,-infinity,",",-8-6*sqrt(2))),
line(-0.7,1,-0.7,-3),
line(-0.7,1,-1,1),
line(-0.7,-3,-1,-3),
locate(0,0,"U"),

line(0.7,1,0.7,-3),
line(0.7,1,1,1),
line(0.7,-3,1,-3),
locate(1,0,matrix(1,3,-8+6*sqrt(2),",",infinity)),
arc(2.3,-1,3,5,180+140,220+180)
)
}}}
The square brackets indicate we include both {{{y=-8-6*sqrt(2)}}} and {{{y=-8+6*sqrt(2)}}} in the range.
Curved parenthesis are <u>always</u> with either infinity. 
This is because we cannot reach infinity.
It's not on the number line. Rather it's a concept that represents going on forever in some direction.


If you prefer decimal approximations, then,
{{{y=-8-6*sqrt(2) = -16.485}}} and {{{y=-8+6*sqrt(2) = 0.485}}} when rounding to 3 decimal places.
The interval notation would then become (-infinity, -16.485] U [0.485, infinity)
The "U" symbol represents "set union". Informally it means "or", so the possible range of values is the interval (-infinity, -16.485] OR it's the interval [0.485, infinity)
Replace the word "infinity" with the infinity symbol if necessary.
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