How do you find domain and range for $f(x)=(x+4)/(x^2-4)$ ?

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Answer 1 · 2017-11-29T17:45:08

The domain is $RR-{-2,2}$ .
The range is $y in (-oo,-0.93] uu [-0.67,+oo)$

As we cannot divide by $0$ , the denominator is $!=0$

$x^2-4=(x+2)(x-2)!=0$

Therefore,

The domain is $RR-{-2,2}$

To determine the range, proceed as follows

Let, $y=(x+4)/(x^2-4)$

$y(x^2-4)=x+4$

$yx^2-x-4(y+4)=0$

This is a quadratic equation in $x^2$ and in order for this equation to have solutions, the discriminant $Delta>=0$

$Delta=b^2-4ac=(-1)^2-4(y)(-4(y+4))>=0$

$1+16y^2+16y>=0$

$16y^2+16y+1>=0$

$y=(-16+-sqrt(16^2-4*16))/(2*16)=(-16+-sqrt(192))/(32)$

$y_1=(-16+sqrt192)/(32)=-0.67$

$y_2=(-16-sqrt192)/(32)=-0.93$

Therefore,

The range is $y in (-oo,-0.93] uu [-0.67,+oo)$

graph{(x+4)/(x^2-4) [-8.89, 8.89, -4.444, 4.445]}

How do you find domain and range for f(x)=(x+4)/(x^2-4) f(x)=(x+4)/(x^2-4) ?