Answer:
![-2 \le x < 5](https://img.qammunity.org/2023/formulas/mathematics/high-school/zalyr319y6tdhc1n88peefxnct7m2nv89o.png)
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Step-by-step explanation:
We cannot have negative numbers under the square root, if we wanted the result to be some real number.
The stuff under the square root must be 0 or larger.
This means the x+2 must be 0 or larger.
![x+2 \ge 0\\\\x \ge -2](https://img.qammunity.org/2023/formulas/mathematics/high-school/yc76vd2ijprpv5qd4xk70h31vw12238r56.png)
Similarly, the 5-x must be 0 or larger. But wait, we cannot have 0 in the denominator (or else we have a division by zero error), so 5-x must be larger than 0.
5-x > 0
5-x+x > 0+x
5+0 > 0+x
5 > x
x < 5
Combine both
and
to find the domain is
![-2 \le x < 5](https://img.qammunity.org/2023/formulas/mathematics/high-school/zalyr319y6tdhc1n88peefxnct7m2nv89o.png)
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Visual confirmation is shown below. I used Desmos which is a free graphing app.
- f(x) = sqrt(x+2) is in red
- g(x) = sqrt(5-x) is in blue
- h(x) = f(x)/g(x) is in green
The green curve is what we're after. It's between x = -2 and x = 5
We include -2, but exclude 5.
Take note of the closed endpoint at x = -2, and also the vertical asymptote at x = 5. The curve approaches this asymptote but never actually touches it. Think of an electric fence you can get closer to, but not actually touch.