How do you find the domain and range for $y = \sqrt{9 - x^{2}}$ $y =sqrt(9-x^2)$ ?

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How do you find the domain and range for $y = \sqrt{9 - x^{2}}$ $y =sqrt(9-x^2)$ ?

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Answer 1 · 2016-02-18T11:54:09

Domain: $x \in [- 3, + 3]$ $x in [-3,+3]$
Range: $y \in [0, 3]$ $yin [0,3]$

Explanation:

I assume that we are dealing only with Real numbers.

For Real numbers $\sqrt{a}$ $sqrt(a)$ is only defined if $a \geq 0$ $a>=0$
$\to (9 - x^{2}) \geq 0$ $rarr (9-x^2)>=0$
$\to | x | \leq 3$ $rarr abs(x)<=3$
which gives us the Domain.

The maximum value of $\sqrt{9 - x^{2}}$ $sqrt(9-x^2)$ occurs when $x = 0$ $x=0$ and
this maximum is $\sqrt{9} = 3$ $sqrt(9) = 3$
The minimum value of $\sqrt{9 - x^{2}}$ $sqrt(9-x^2)$ occurs when $x = \pm 3$ $x=+-3$ and
this minimum is $\sqrt{9 - 9} = 0$ $sqrt(9-9) =0$
which together give us the Range.

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How do you find the domain and range for $y = \sqrt{9 - x^{2}}$ $y =sqrt(9-x^2)$ ?

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Explanation:

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How do you find the domain and range for y=√9−x2y =sqrt(9-x^2)?

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How do you find the domain and range for $y = \sqrt{9 - x^{2}}$ $y =sqrt(9-x^2)$ ?