Find the largest possible domain for the real valued function given by $f (x) = \frac{\sqrt{9 - x^{2}}}{\sqrt{x^{2} - 1}} .$

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Solution

Given

$\frac{\sqrt{9 - x^{2}}}{\sqrt{x^{2} - 1}}$

Domain of $9 - x^{2}$

$9 - x^{2} \geq 0$

$9 \geq x^{2}$

$x^{2} \leq 9$

$x \leq \pm 3$

Domain of $\sqrt{x^{2} - 1}$

$x^{2} - 1 >$

$x^{2} > 1$

$x > \pm 1$

$∴ D o m a i n o f \frac{\sqrt{9 - x^{2}}}{x^{2} - 1} i s - 1 < x \leq 3$

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Find the largest possible domain for the real valued function given by f(x)=√9−x2√x2−1.

Given√9−x2√x2−1Domain of 9−x29−x2≥09≥x2x2≤9x≤±3Domain of √x2−1x2−1>x2>1x>±1∴Domainof√9−x2x2−1is−1<x≤3

Find the largest possible domain for the real valued function given by $f (x) = \frac{\sqrt{9 - x^{2}}}{\sqrt{x^{2} - 1}} .$

Given

$\frac{\sqrt{9 - x^{2}}}{\sqrt{x^{2} - 1}}$

Domain of $9 - x^{2}$

$9 - x^{2} \geq 0$

$9 \geq x^{2}$

$x^{2} \leq 9$

$x \leq \pm 3$

Domain of $\sqrt{x^{2} - 1}$

$x^{2} - 1 >$

$x^{2} > 1$

$x > \pm 1$

$∴ D o m a i n o f \frac{\sqrt{9 - x^{2}}}{x^{2} - 1} i s - 1 < x \leq 3$