How do you find the range of the function $y=f(x)=x/(x^2-5x+9)$ ?

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Answer 1 · 2015-04-22T19:15:33

The answer is : $-1/11<=y<=1$

Solution
$y=x/(x^2-5x+9)$

$=>yx^2-5yx+9y=x$

Send all terms to the left

$=> yx^2+(-1-5y)x+9y=0$

For real roots, $b^2-4ac>=0$

$=> (-1-5y)^2-4(y)(9y)>=0$

$=> 1+10y-11y^2>=0$

Now, all you've got to do is to factorize and find the range :)

$=>(-1-11y)(-1+y)>=0$

Or $(11y+1)(y-1)<=0$

Hence $-1/11<=y<=1$

How do you find the range of the function y=f(x)=x/(x^2-5x+9)y=f(x)=x/(x^2-5x+9)?