What is the range of the function $h (x) = \frac{1}{x^{2} - 9}$ $h(x) = 1 /(x^2 - 9)$ ?

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What is the range of the function $h (x) = \frac{1}{x^{2} - 9}$ $h(x) = 1 /(x^2 - 9)$ ?

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Answer 1 · 2017-04-17T07:33:07

$y \geq - \frac{1}{9}$ $y\geq -1/9$ , $y \neq 0$ $yne 0$

Explanation:

$y$ $y$ is in the range if and only if the polynomial $p (x) = y x^{2} - y 9 - 1 = 0$ $p(x)=yx^2-y9-1=0$ has a root, and it has a root if and only if $9 + \frac{1}{y} \geq 0$ $9+1/ygeq 0$ and $y \neq 0$ $yne 0$ , so
$y \geq - \frac{1}{9}$ $y\geq -1/9$ , $y \neq 0$ $yne 0$

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What is the range of the function $h (x) = \frac{1}{x^{2} - 9}$ $h(x) = 1 /(x^2 - 9)$ ?

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What is the range of the function h(x) = 1 /(x^2 - 9)h(x)=1x2−9 h(x) = 1 /(x^2 - 9)?

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What is the range of the function $h (x) = \frac{1}{x^{2} - 9}$ $h(x) = 1 /(x^2 - 9)$ ?