How do you find the domain and range of $y = 5 - \sqrt{9 - x^{2}}$ $y = 5 - sqrt(9-x^2)$ ?

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How do you find the domain and range of $y = 5 - \sqrt{9 - x^{2}}$ $y = 5 - sqrt(9-x^2)$ ?

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Answer 1 · 2018-07-22T05:29:40

$D x = [- 3; 3] and R = [2; 5]$ $Dx=[-3;3] and R=[2;5]$

Explanation:

$y = 5 - \sqrt{9 - x^{2}}$ $y=5-sqrt(9-x^2)$

domain $\sqrt{m}$ $sqrt(m)$ then $d o m = m \geq 0$ $dom=m>=0$

$\sqrt{9 - x^{2}} \geq 0$ $sqrt(9-x^2)>=0$

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

$x^{2} - 9 \leq 0$ $x^2-9<=0$

$(x - 3) (x + 3) \leq 0$ $(x-3)(x+3)<=0$

$x = - 3; x = 3$ $x=-3; x=3$

negative zone

$- 3 \leq x \leq 3 \to x \in [- 3; 3]$ $-3<=x<=3 -> x ∈ [-3;3]$

Range from domain

$- 3 \leq x \leq 3$ $-3<=x<=3$

elevating to square

$0 \leq x^{2} \leq 9$ $0<=x^2<=9$

multiply -1

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

add 9

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

square root

$3 \geq \sqrt{9 - x^{2}} \geq 0$ $3>=sqrt(9-x^2)>=0$

multiply -1

$- 3 \leq - \sqrt{9 - x^{2}} \leq 0$ $-3<=-sqrt(9-x^2)<=0$

add 5

$2 \leq 5 - \sqrt{9 - x^{2}} \leq 5$ $2<=5-sqrt(9-x^2)<=5$

$2 \leq y \leq 5$ $2<=y<=5$

Range $y \in [2; 5]$ $y ∈ [2;5]$

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How do you find the domain and range of $y = 5 - \sqrt{9 - x^{2}}$ $y = 5 - sqrt(9-x^2)$ ?

1 Answer

Explanation:

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How do you find the domain and range of y=5−√9−x2 y = 5 - sqrt(9-x^2)?

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Explanation:

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How do you find the domain and range of $y = 5 - \sqrt{9 - x^{2}}$ $y = 5 - sqrt(9-x^2)$ ?