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

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Answer 1 · 2015-04-22T19:15:33

The answer is : $- \frac{1}{11} \leq y \leq 1$ $-1/11<=y<=1$

Solution
$y = \frac{x}{x^{2} - 5 x + 9}$ $y=x/(x^2-5x+9)$

$\Rightarrow y x^{2} - 5 y x + 9 y = x$ $=>yx^2-5yx+9y=x$

Send all terms to the left

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

For real roots, $b^{2} - 4 a c \geq 0$ $b^2-4ac>=0$

$\Rightarrow {(- 1 - 5 y)}^{2} - 4 (y) (9 y) \geq 0$ $=> (-1-5y)^2-4(y)(9y)>=0$

$\Rightarrow 1 + 10 y - 11 y^{2} \geq 0$ $=> 1+10y-11y^2>=0$

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

$\Rightarrow (- 1 - 11 y) (- 1 + y) \geq 0$ $=>(-1-11y)(-1+y)>=0$

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

Hence $- \frac{1}{11} \leq y \leq 1$ $-1/11<=y<=1$

How do you find the range of the function y=f(x)=xx2−5x+9y=f(x)=x/(x^2-5x+9)?