How do you find the domain and range of $f(x) = (2x+1) / (2x^2 + 5x + 2)$ ?

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Answer 1 · 2017-09-08T08:38:40

The domain is $x in RR//{-2}$
The range is $y in RR//{0}$

The denominator must de $!=0$

So,

$2x^2+5x+2 = (2x+1)(x+2)!=0$

$f(x)=(2x+1)/((2x+1)(x+2))=1/(x+2)$

Therefore,

The domain is $x in RR//{-2}$

To find the range, we proceed as follows

Let $y=(2x+1)/(2x^2+5x+2 )=1/(x+2)$

$y(x+2 )=1$

$yx+2y=1$

$x=(1-2y)/y$

The denominator must be $!=0$

Therefore,

The range is $y in RR//{0}$

graph{(2x+1)/(2x^2+5x+2) [-18.02, 18.03, -9.01, 9.01]}

How do you find the domain and range of f(x) = (2x+1) / (2x^2 + 5x + 2)f(x) = (2x+1) / (2x^2 + 5x + 2)?