Domain — Quotient of Two Square Roots — Dynamic Practice

Domain — Quotient of Two Square Roots — Dynamic Practice. Practice questions to deepen understanding of the domain when a quotient of two square roots appears. Online dynamic learning math system.

Dynamic practice in the domain of a quotient of two square roots — numerator radicand must be non-negative AND denominator radicand must be strictly positive.

42 questions

Question 1

2.38 pts

Find the domain of the function:

\(f(x) = \frac{\sqrt{x-2}}{\sqrt{x}}\)

\(x \geq 2\)

\(x > 2\)

\(x > 0\)

\(x \neq 0\)

Explanation:

Solution — Domain:

The function: \(f(x) = \frac{\sqrt{x-2}}{\sqrt{x}}\)

📌 Step 1: Square root in numerator — the expression must be non-negative

\(x - 2 \geq 0 \;\Rightarrow\; x \geq 2\)

📌 Step 2: Square root in denominator — the expression must be strictly positive

\(x - 0 > 0 \;\Rightarrow\; x > 0\)

📌 Step 3: Intersection of the two conditions

\(x \geq 2\) and \(x > 0\)

Take the stricter: \(x \geq 2\)

Domain: \(x \geq 2\)

Question 2

2.38 pts

Find the domain of the function:

\(f(x) = \frac{\sqrt{x-6}}{\sqrt{x-1}}\)

\(x \geq 6\)

\(x > 6\)

\(x > 1\)

\(x \neq 1\)

Explanation:

Solution — Domain:

The function: \(f(x) = \frac{\sqrt{x-6}}{\sqrt{x-1}}\)

📌 Step 1: Square root in numerator — the expression must be non-negative

\(x - 6 \geq 0 \;\Rightarrow\; x \geq 6\)

📌 Step 2: Square root in denominator — the expression must be strictly positive

\(x - 1 > 0 \;\Rightarrow\; x > 1\)

📌 Step 3: Intersection of the two conditions

\(x \geq 6\) and \(x > 1\)

Take the stricter: \(x \geq 6\)

Domain: \(x \geq 6\)

Question 3

2.38 pts

Find the domain of the function:

\(f(x) = \frac{\sqrt{x-9}}{\sqrt{x-2}}\)

\(x \geq 9\)

\(x > 9\)

\(x > 2\)

\(x \neq 2\)

Explanation:

Solution — Domain:

The function: \(f(x) = \frac{\sqrt{x-9}}{\sqrt{x-2}}\)

📌 Step 1: Square root in numerator — the expression must be non-negative

\(x - 9 \geq 0 \;\Rightarrow\; x \geq 9\)

📌 Step 2: Square root in denominator — the expression must be strictly positive

\(x - 2 > 0 \;\Rightarrow\; x > 2\)

📌 Step 3: Intersection of the two conditions

\(x \geq 9\) and \(x > 2\)

Take the stricter: \(x \geq 9\)

Domain: \(x \geq 9\)

Question 4

2.38 pts

Find the domain of the function:

\(f(x) = \frac{\sqrt{x}}{\sqrt{x+3}}\)

\(x \geq 0\)

\(x > 0\)

\(x > (-3)\)

\(x \neq (-3)\)

Explanation:

Solution — Domain:

The function: \(f(x) = \frac{\sqrt{x}}{\sqrt{x+3}}\)

📌 Step 1: Square root in numerator — the expression must be non-negative

\(x - 0 \geq 0 \;\Rightarrow\; x \geq 0\)

📌 Step 2: Square root in denominator — the expression must be strictly positive

\(x - -3 > 0 \;\Rightarrow\; x > -3\)

📌 Step 3: Intersection of the two conditions

\(x \geq 0\) and \(x > -3\)

Take the stricter: \(x \geq 0\)

Domain: \(x \geq 0\)

Question 5

2.38 pts

Find the domain of the function:

\(f(x) = \frac{\sqrt{x+4}}{\sqrt{x-4}}\)

\(x > 4\)

\(x \geq 4\)

\(x > (-4)\)

\(x \neq 4\)

Explanation:

Solution — Domain:

The function: \(f(x) = \frac{\sqrt{x+4}}{\sqrt{x-4}}\)

📌 Step 1: Square root in numerator — the expression must be non-negative

\(x - -4 \geq 0 \;\Rightarrow\; x \geq -4\)

📌 Step 2: Square root in denominator — the expression must be strictly positive

\(x - 4 > 0 \;\Rightarrow\; x > 4\)

📌 Step 3: Intersection of the two conditions

\(x \geq -4\) and \(x > 4\)

Take the stricter: \(x > 4\)

Domain: \(x > 4\)

Question 6

2.38 pts

Find the domain of the function:

\(f(x) = \frac{\sqrt{x-6}}{\sqrt{x+3}}\)

\(x \geq 6\)

\(x > 6\)

\(x > (-3)\)

\(x \neq (-3)\)

Explanation:

Solution — Domain:

The function: \(f(x) = \frac{\sqrt{x-6}}{\sqrt{x+3}}\)

📌 Step 1: Square root in numerator — the expression must be non-negative

\(x - 6 \geq 0 \;\Rightarrow\; x \geq 6\)

📌 Step 2: Square root in denominator — the expression must be strictly positive

\(x - -3 > 0 \;\Rightarrow\; x > -3\)

📌 Step 3: Intersection of the two conditions

\(x \geq 6\) and \(x > -3\)

Take the stricter: \(x \geq 6\)

Domain: \(x \geq 6\)

Question 7

2.38 pts

Find the domain of the function:

\(f(x) = \frac{\sqrt{x-7}}{\sqrt{x+3}}\)

\(x \geq 7\)

\(x > 7\)

\(x > (-3)\)

\(x \neq (-3)\)

Explanation:

Solution — Domain:

The function: \(f(x) = \frac{\sqrt{x-7}}{\sqrt{x+3}}\)

📌 Step 1: Square root in numerator — the expression must be non-negative

\(x - 7 \geq 0 \;\Rightarrow\; x \geq 7\)

📌 Step 2: Square root in denominator — the expression must be strictly positive

\(x - -3 > 0 \;\Rightarrow\; x > -3\)

📌 Step 3: Intersection of the two conditions

\(x \geq 7\) and \(x > -3\)

Take the stricter: \(x \geq 7\)

Domain: \(x \geq 7\)

Question 8

2.38 pts

Find the domain of the function:

\(f(x) = \frac{\sqrt{x-8}}{\sqrt{x-7}}\)

\(x \geq 8\)

\(x > 8\)

\(x > 7\)

\(x \neq 7\)

Explanation:

Solution — Domain:

The function: \(f(x) = \frac{\sqrt{x-8}}{\sqrt{x-7}}\)

📌 Step 1: Square root in numerator — the expression must be non-negative

\(x - 8 \geq 0 \;\Rightarrow\; x \geq 8\)

📌 Step 2: Square root in denominator — the expression must be strictly positive

\(x - 7 > 0 \;\Rightarrow\; x > 7\)

📌 Step 3: Intersection of the two conditions

\(x \geq 8\) and \(x > 7\)

Take the stricter: \(x \geq 8\)

Domain: \(x \geq 8\)

Question 9

2.38 pts

Find the domain of the function:

\(f(x) = \frac{\sqrt{x-6}}{\sqrt{x+6}}\)

\(x \geq 6\)

\(x > 6\)

\(x > (-6)\)

\(x \neq (-6)\)

Explanation:

Solution — Domain:

The function: \(f(x) = \frac{\sqrt{x-6}}{\sqrt{x+6}}\)

📌 Step 1: Square root in numerator — the expression must be non-negative

\(x - 6 \geq 0 \;\Rightarrow\; x \geq 6\)

📌 Step 2: Square root in denominator — the expression must be strictly positive

\(x - -6 > 0 \;\Rightarrow\; x > -6\)

📌 Step 3: Intersection of the two conditions

\(x \geq 6\) and \(x > -6\)

Take the stricter: \(x \geq 6\)

Domain: \(x \geq 6\)

Question 10

2.38 pts

Find the domain of the function:

\(f(x) = \frac{\sqrt{x+4}}{\sqrt{x-2}}\)

\(x > 2\)

\(x \geq 2\)

\(x > (-4)\)

\(x \neq 2\)

Explanation:

Solution — Domain:

The function: \(f(x) = \frac{\sqrt{x+4}}{\sqrt{x-2}}\)

📌 Step 1: Square root in numerator — the expression must be non-negative

\(x - -4 \geq 0 \;\Rightarrow\; x \geq -4\)

📌 Step 2: Square root in denominator — the expression must be strictly positive

\(x - 2 > 0 \;\Rightarrow\; x > 2\)

📌 Step 3: Intersection of the two conditions

\(x \geq -4\) and \(x > 2\)

Take the stricter: \(x > 2\)

Domain: \(x > 2\)

Question 11

2.38 pts

Find the domain of the function:

\(f(x) = \frac{\sqrt{x-5}}{\sqrt{x-6}}\)

\(x > 6\)

\(x \geq 6\)

\(x > 5\)

\(x \neq 6\)

Explanation:

Solution — Domain:

The function: \(f(x) = \frac{\sqrt{x-5}}{\sqrt{x-6}}\)

📌 Step 1: Square root in numerator — the expression must be non-negative

\(x - 5 \geq 0 \;\Rightarrow\; x \geq 5\)

📌 Step 2: Square root in denominator — the expression must be strictly positive

\(x - 6 > 0 \;\Rightarrow\; x > 6\)

📌 Step 3: Intersection of the two conditions

\(x \geq 5\) and \(x > 6\)

Take the stricter: \(x > 6\)

Domain: \(x > 6\)

Question 12

2.38 pts

Find the domain of the function:

\(f(x) = \frac{\sqrt{x+1}}{\sqrt{x+2}}\)

\(x \geq (-1)\)

\(x > (-1)\)

\(x > (-2)\)

\(x \neq (-2)\)

Explanation:

Solution — Domain:

The function: \(f(x) = \frac{\sqrt{x+1}}{\sqrt{x+2}}\)

📌 Step 1: Square root in numerator — the expression must be non-negative

\(x - -1 \geq 0 \;\Rightarrow\; x \geq -1\)

📌 Step 2: Square root in denominator — the expression must be strictly positive

\(x - -2 > 0 \;\Rightarrow\; x > -2\)

📌 Step 3: Intersection of the two conditions

\(x \geq -1\) and \(x > -2\)

Take the stricter: \(x \geq -1\)

Domain: \(x \geq (-1)\)

Question 13

2.38 pts

Find the domain of the function:

\(f(x) = \frac{\sqrt{x-7}}{\sqrt{x+6}}\)

\(x \geq 7\)

\(x > 7\)

\(x > (-6)\)

\(x \neq (-6)\)

Explanation:

Solution — Domain:

The function: \(f(x) = \frac{\sqrt{x-7}}{\sqrt{x+6}}\)

📌 Step 1: Square root in numerator — the expression must be non-negative

\(x - 7 \geq 0 \;\Rightarrow\; x \geq 7\)

📌 Step 2: Square root in denominator — the expression must be strictly positive

\(x - -6 > 0 \;\Rightarrow\; x > -6\)

📌 Step 3: Intersection of the two conditions

\(x \geq 7\) and \(x > -6\)

Take the stricter: \(x \geq 7\)

Domain: \(x \geq 7\)

Question 14

2.38 pts

Find the domain of the function:

\(f(x) = \frac{\sqrt{x+7}}{\sqrt{x+1}}\)

\(x > (-1)\)

\(x \geq (-1)\)

\(x > (-7)\)

\(x \neq (-1)\)

Explanation:

Solution — Domain:

The function: \(f(x) = \frac{\sqrt{x+7}}{\sqrt{x+1}}\)

📌 Step 1: Square root in numerator — the expression must be non-negative

\(x - -7 \geq 0 \;\Rightarrow\; x \geq -7\)

📌 Step 2: Square root in denominator — the expression must be strictly positive

\(x - -1 > 0 \;\Rightarrow\; x > -1\)

📌 Step 3: Intersection of the two conditions

\(x \geq -7\) and \(x > -1\)

Take the stricter: \(x > -1\)

Domain: \(x > (-1)\)

Question 15

2.38 pts

Find the domain of the function:

\(f(x) = \frac{\sqrt{x+4}}{\sqrt{x-8}}\)

\(x > 8\)

\(x \geq 8\)

\(x > (-4)\)

\(x \neq 8\)

Explanation:

Solution — Domain:

The function: \(f(x) = \frac{\sqrt{x+4}}{\sqrt{x-8}}\)

📌 Step 1: Square root in numerator — the expression must be non-negative

\(x - -4 \geq 0 \;\Rightarrow\; x \geq -4\)

📌 Step 2: Square root in denominator — the expression must be strictly positive

\(x - 8 > 0 \;\Rightarrow\; x > 8\)

📌 Step 3: Intersection of the two conditions

\(x \geq -4\) and \(x > 8\)

Take the stricter: \(x > 8\)

Domain: \(x > 8\)

Question 16

2.38 pts

Find the domain of the function:

\(f(x) = \frac{\sqrt{x-1}}{\sqrt{x+2}}\)

\(x \geq 1\)

\(x > 1\)

\(x > (-2)\)

\(x \neq (-2)\)

Explanation:

Solution — Domain:

The function: \(f(x) = \frac{\sqrt{x-1}}{\sqrt{x+2}}\)

📌 Step 1: Square root in numerator — the expression must be non-negative

\(x - 1 \geq 0 \;\Rightarrow\; x \geq 1\)

📌 Step 2: Square root in denominator — the expression must be strictly positive

\(x - -2 > 0 \;\Rightarrow\; x > -2\)

📌 Step 3: Intersection of the two conditions

\(x \geq 1\) and \(x > -2\)

Take the stricter: \(x \geq 1\)

Domain: \(x \geq 1\)

Question 17

2.38 pts

Find the domain of the function:

\(f(x) = \frac{\sqrt{x+9}}{\sqrt{x-6}}\)

\(x > 6\)

\(x \geq 6\)

\(x > (-9)\)

\(x \neq 6\)

Explanation:

Solution — Domain:

The function: \(f(x) = \frac{\sqrt{x+9}}{\sqrt{x-6}}\)

📌 Step 1: Square root in numerator — the expression must be non-negative

\(x - -9 \geq 0 \;\Rightarrow\; x \geq -9\)

📌 Step 2: Square root in denominator — the expression must be strictly positive

\(x - 6 > 0 \;\Rightarrow\; x > 6\)

📌 Step 3: Intersection of the two conditions

\(x \geq -9\) and \(x > 6\)

Take the stricter: \(x > 6\)

Domain: \(x > 6\)

Question 18

2.38 pts

Find the domain of the function:

\(f(x) = \frac{\sqrt{x}}{\sqrt{x-3}}\)

\(x > 3\)

\(x \geq 3\)

\(x > 0\)

\(x \neq 3\)

Explanation:

Solution — Domain:

The function: \(f(x) = \frac{\sqrt{x}}{\sqrt{x-3}}\)

📌 Step 1: Square root in numerator — the expression must be non-negative

\(x - 0 \geq 0 \;\Rightarrow\; x \geq 0\)

📌 Step 2: Square root in denominator — the expression must be strictly positive

\(x - 3 > 0 \;\Rightarrow\; x > 3\)

📌 Step 3: Intersection of the two conditions

\(x \geq 0\) and \(x > 3\)

Take the stricter: \(x > 3\)

Domain: \(x > 3\)

Question 19

2.38 pts

Find the domain of the function:

\(f(x) = \frac{\sqrt{x-9}}{\sqrt{x+5}}\)

\(x \geq 9\)

\(x > 9\)

\(x > (-5)\)

\(x \neq (-5)\)

Explanation:

Solution — Domain:

The function: \(f(x) = \frac{\sqrt{x-9}}{\sqrt{x+5}}\)

📌 Step 1: Square root in numerator — the expression must be non-negative

\(x - 9 \geq 0 \;\Rightarrow\; x \geq 9\)

📌 Step 2: Square root in denominator — the expression must be strictly positive

\(x - -5 > 0 \;\Rightarrow\; x > -5\)

📌 Step 3: Intersection of the two conditions

\(x \geq 9\) and \(x > -5\)

Take the stricter: \(x \geq 9\)

Domain: \(x \geq 9\)

Question 20

2.38 pts

Find the domain of the function:

\(f(x) = \frac{\sqrt{x+3}}{\sqrt{x-9}}\)

\(x > 9\)

\(x \geq 9\)

\(x > (-3)\)

\(x \neq 9\)

Explanation:

Solution — Domain:

The function: \(f(x) = \frac{\sqrt{x+3}}{\sqrt{x-9}}\)

📌 Step 1: Square root in numerator — the expression must be non-negative

\(x - -3 \geq 0 \;\Rightarrow\; x \geq -3\)

📌 Step 2: Square root in denominator — the expression must be strictly positive

\(x - 9 > 0 \;\Rightarrow\; x > 9\)

📌 Step 3: Intersection of the two conditions

\(x \geq -3\) and \(x > 9\)

Take the stricter: \(x > 9\)

Domain: \(x > 9\)

Question 21

2.38 pts

Find the domain of the function:

\(f(x) = \frac{\sqrt{x-1}}{\sqrt{x+5}}\)

\(x \geq 1\)

\(x > 1\)

\(x > (-5)\)

\(x \neq (-5)\)

Explanation:

Solution — Domain:

The function: \(f(x) = \frac{\sqrt{x-1}}{\sqrt{x+5}}\)

📌 Step 1: Square root in numerator — the expression must be non-negative

\(x - 1 \geq 0 \;\Rightarrow\; x \geq 1\)

📌 Step 2: Square root in denominator — the expression must be strictly positive

\(x - -5 > 0 \;\Rightarrow\; x > -5\)

📌 Step 3: Intersection of the two conditions

\(x \geq 1\) and \(x > -5\)

Take the stricter: \(x \geq 1\)

Domain: \(x \geq 1\)

Question 22

2.38 pts

Find the domain of the function:

\(f(x) = \frac{\sqrt{x-2}}{\sqrt{x+1}}\)

\(x \geq 2\)

\(x > 2\)

\(x > (-1)\)

\(x \neq (-1)\)

Explanation:

Solution — Domain:

The function: \(f(x) = \frac{\sqrt{x-2}}{\sqrt{x+1}}\)

📌 Step 1: Square root in numerator — the expression must be non-negative

\(x - 2 \geq 0 \;\Rightarrow\; x \geq 2\)

📌 Step 2: Square root in denominator — the expression must be strictly positive

\(x - -1 > 0 \;\Rightarrow\; x > -1\)

📌 Step 3: Intersection of the two conditions

\(x \geq 2\) and \(x > -1\)

Take the stricter: \(x \geq 2\)

Domain: \(x \geq 2\)

Question 23

2.38 pts

Find the domain of the function:

\(f(x) = \frac{\sqrt{x-8}}{\sqrt{x+2}}\)

\(x \geq 8\)

\(x > 8\)

\(x > (-2)\)

\(x \neq (-2)\)

Explanation:

Solution — Domain:

The function: \(f(x) = \frac{\sqrt{x-8}}{\sqrt{x+2}}\)

📌 Step 1: Square root in numerator — the expression must be non-negative

\(x - 8 \geq 0 \;\Rightarrow\; x \geq 8\)

📌 Step 2: Square root in denominator — the expression must be strictly positive

\(x - -2 > 0 \;\Rightarrow\; x > -2\)

📌 Step 3: Intersection of the two conditions

\(x \geq 8\) and \(x > -2\)

Take the stricter: \(x \geq 8\)

Domain: \(x \geq 8\)

Question 24

2.38 pts

Find the domain of the function:

\(f(x) = \frac{\sqrt{x+2}}{\sqrt{x+1}}\)

\(x > (-1)\)

\(x \geq (-1)\)

\(x > (-2)\)

\(x \neq (-1)\)

Explanation:

Solution — Domain:

The function: \(f(x) = \frac{\sqrt{x+2}}{\sqrt{x+1}}\)

📌 Step 1: Square root in numerator — the expression must be non-negative

\(x - -2 \geq 0 \;\Rightarrow\; x \geq -2\)

📌 Step 2: Square root in denominator — the expression must be strictly positive

\(x - -1 > 0 \;\Rightarrow\; x > -1\)

📌 Step 3: Intersection of the two conditions

\(x \geq -2\) and \(x > -1\)

Take the stricter: \(x > -1\)

Domain: \(x > (-1)\)

Question 25

2.38 pts

Find the domain of the function:

\(f(x) = \frac{\sqrt{x-6}}{\sqrt{x-2}}\)

\(x \geq 6\)

\(x > 6\)

\(x > 2\)

\(x \neq 2\)

Explanation:

Solution — Domain:

The function: \(f(x) = \frac{\sqrt{x-6}}{\sqrt{x-2}}\)

📌 Step 1: Square root in numerator — the expression must be non-negative

\(x - 6 \geq 0 \;\Rightarrow\; x \geq 6\)

📌 Step 2: Square root in denominator — the expression must be strictly positive

\(x - 2 > 0 \;\Rightarrow\; x > 2\)

📌 Step 3: Intersection of the two conditions

\(x \geq 6\) and \(x > 2\)

Take the stricter: \(x \geq 6\)

Domain: \(x \geq 6\)

Question 26

2.38 pts

Find the domain of the function:

\(f(x) = \frac{\sqrt{x-7}}{\sqrt{x-2}}\)

\(x \geq 7\)

\(x > 7\)

\(x > 2\)

\(x \neq 2\)

Explanation:

Solution — Domain:

The function: \(f(x) = \frac{\sqrt{x-7}}{\sqrt{x-2}}\)

📌 Step 1: Square root in numerator — the expression must be non-negative

\(x - 7 \geq 0 \;\Rightarrow\; x \geq 7\)

📌 Step 2: Square root in denominator — the expression must be strictly positive

\(x - 2 > 0 \;\Rightarrow\; x > 2\)

📌 Step 3: Intersection of the two conditions

\(x \geq 7\) and \(x > 2\)

Take the stricter: \(x \geq 7\)

Domain: \(x \geq 7\)

Question 27

2.38 pts

Find the domain of the function:

\(f(x) = \frac{\sqrt{x+4}}{\sqrt{x+9}}\)

\(x \geq (-4)\)

\(x > (-4)\)

\(x > (-9)\)

\(x \neq (-9)\)

Explanation:

Solution — Domain:

The function: \(f(x) = \frac{\sqrt{x+4}}{\sqrt{x+9}}\)

📌 Step 1: Square root in numerator — the expression must be non-negative

\(x - -4 \geq 0 \;\Rightarrow\; x \geq -4\)

📌 Step 2: Square root in denominator — the expression must be strictly positive

\(x - -9 > 0 \;\Rightarrow\; x > -9\)

📌 Step 3: Intersection of the two conditions

\(x \geq -4\) and \(x > -9\)

Take the stricter: \(x \geq -4\)

Domain: \(x \geq (-4)\)

Question 28

2.38 pts

Find the domain of the function:

\(f(x) = \frac{\sqrt{x+6}}{\sqrt{x-4}}\)

\(x > 4\)

\(x \geq 4\)

\(x > (-6)\)

\(x \neq 4\)

Explanation:

Solution — Domain:

The function: \(f(x) = \frac{\sqrt{x+6}}{\sqrt{x-4}}\)

📌 Step 1: Square root in numerator — the expression must be non-negative

\(x - -6 \geq 0 \;\Rightarrow\; x \geq -6\)

📌 Step 2: Square root in denominator — the expression must be strictly positive

\(x - 4 > 0 \;\Rightarrow\; x > 4\)

📌 Step 3: Intersection of the two conditions

\(x \geq -6\) and \(x > 4\)

Take the stricter: \(x > 4\)

Domain: \(x > 4\)

Question 29

2.38 pts

Find the domain of the function:

\(f(x) = \frac{\sqrt{x+7}}{\sqrt{x-9}}\)

\(x > 9\)

\(x \geq 9\)

\(x > (-7)\)

\(x \neq 9\)

Explanation:

Solution — Domain:

The function: \(f(x) = \frac{\sqrt{x+7}}{\sqrt{x-9}}\)

📌 Step 1: Square root in numerator — the expression must be non-negative

\(x - -7 \geq 0 \;\Rightarrow\; x \geq -7\)

📌 Step 2: Square root in denominator — the expression must be strictly positive

\(x - 9 > 0 \;\Rightarrow\; x > 9\)

📌 Step 3: Intersection of the two conditions

\(x \geq -7\) and \(x > 9\)

Take the stricter: \(x > 9\)

Domain: \(x > 9\)

Question 30

2.38 pts

Find the domain of the function:

\(f(x) = \frac{\sqrt{x+8}}{\sqrt{x+3}}\)

\(x > (-3)\)

\(x \geq (-3)\)

\(x > (-8)\)

\(x \neq (-3)\)

Explanation:

Solution — Domain:

The function: \(f(x) = \frac{\sqrt{x+8}}{\sqrt{x+3}}\)

📌 Step 1: Square root in numerator — the expression must be non-negative

\(x - -8 \geq 0 \;\Rightarrow\; x \geq -8\)

📌 Step 2: Square root in denominator — the expression must be strictly positive

\(x - -3 > 0 \;\Rightarrow\; x > -3\)

📌 Step 3: Intersection of the two conditions

\(x \geq -8\) and \(x > -3\)

Take the stricter: \(x > -3\)

Domain: \(x > (-3)\)

Question 31

2.38 pts

Find the domain of the function:

\(f(x) = \frac{\sqrt{x-1}}{\sqrt{x+3}}\)

\(x \geq 1\)

\(x > 1\)

\(x > (-3)\)

\(x \neq (-3)\)

Explanation:

Solution — Domain:

The function: \(f(x) = \frac{\sqrt{x-1}}{\sqrt{x+3}}\)

📌 Step 1: Square root in numerator — the expression must be non-negative

\(x - 1 \geq 0 \;\Rightarrow\; x \geq 1\)

📌 Step 2: Square root in denominator — the expression must be strictly positive

\(x - -3 > 0 \;\Rightarrow\; x > -3\)

📌 Step 3: Intersection of the two conditions

\(x \geq 1\) and \(x > -3\)

Take the stricter: \(x \geq 1\)

Domain: \(x \geq 1\)

Question 32

2.38 pts

Find the domain of the function:

\(f(x) = \frac{\sqrt{x+8}}{\sqrt{x-3}}\)

\(x > 3\)

\(x \geq 3\)

\(x > (-8)\)

\(x \neq 3\)

Explanation:

Solution — Domain:

The function: \(f(x) = \frac{\sqrt{x+8}}{\sqrt{x-3}}\)

📌 Step 1: Square root in numerator — the expression must be non-negative

\(x - -8 \geq 0 \;\Rightarrow\; x \geq -8\)

📌 Step 2: Square root in denominator — the expression must be strictly positive

\(x - 3 > 0 \;\Rightarrow\; x > 3\)

📌 Step 3: Intersection of the two conditions

\(x \geq -8\) and \(x > 3\)

Take the stricter: \(x > 3\)

Domain: \(x > 3\)

Question 33

2.38 pts

Find the domain of the function:

\(f(x) = \frac{\sqrt{x-1}}{\sqrt{x+7}}\)

\(x \geq 1\)

\(x > 1\)

\(x > (-7)\)

\(x \neq (-7)\)

Explanation:

Solution — Domain:

The function: \(f(x) = \frac{\sqrt{x-1}}{\sqrt{x+7}}\)

📌 Step 1: Square root in numerator — the expression must be non-negative

\(x - 1 \geq 0 \;\Rightarrow\; x \geq 1\)

📌 Step 2: Square root in denominator — the expression must be strictly positive

\(x - -7 > 0 \;\Rightarrow\; x > -7\)

📌 Step 3: Intersection of the two conditions

\(x \geq 1\) and \(x > -7\)

Take the stricter: \(x \geq 1\)

Domain: \(x \geq 1\)

Question 34

2.38 pts

Find the domain of the function:

\(f(x) = \frac{\sqrt{x+2}}{\sqrt{x}}\)

\(x > 0\)

\(x \geq 0\)

\(x > (-2)\)

\(x \neq 0\)

Explanation:

Solution — Domain:

The function: \(f(x) = \frac{\sqrt{x+2}}{\sqrt{x}}\)

📌 Step 1: Square root in numerator — the expression must be non-negative

\(x - -2 \geq 0 \;\Rightarrow\; x \geq -2\)

📌 Step 2: Square root in denominator — the expression must be strictly positive

\(x - 0 > 0 \;\Rightarrow\; x > 0\)

📌 Step 3: Intersection of the two conditions

\(x \geq -2\) and \(x > 0\)

Take the stricter: \(x > 0\)

Domain: \(x > 0\)

Question 35

2.38 pts

Find the domain of the function:

\(f(x) = \frac{\sqrt{x+2}}{\sqrt{x+9}}\)

\(x \geq (-2)\)

\(x > (-2)\)

\(x > (-9)\)

\(x \neq (-9)\)

Explanation:

Solution — Domain:

The function: \(f(x) = \frac{\sqrt{x+2}}{\sqrt{x+9}}\)

📌 Step 1: Square root in numerator — the expression must be non-negative

\(x - -2 \geq 0 \;\Rightarrow\; x \geq -2\)

📌 Step 2: Square root in denominator — the expression must be strictly positive

\(x - -9 > 0 \;\Rightarrow\; x > -9\)

📌 Step 3: Intersection of the two conditions

\(x \geq -2\) and \(x > -9\)

Take the stricter: \(x \geq -2\)

Domain: \(x \geq (-2)\)

Question 36

2.38 pts

Find the domain of the function:

\(f(x) = \frac{\sqrt{x+9}}{\sqrt{x-2}}\)

\(x > 2\)

\(x \geq 2\)

\(x > (-9)\)

\(x \neq 2\)

Explanation:

Solution — Domain:

The function: \(f(x) = \frac{\sqrt{x+9}}{\sqrt{x-2}}\)

📌 Step 1: Square root in numerator — the expression must be non-negative

\(x - -9 \geq 0 \;\Rightarrow\; x \geq -9\)

📌 Step 2: Square root in denominator — the expression must be strictly positive

\(x - 2 > 0 \;\Rightarrow\; x > 2\)

📌 Step 3: Intersection of the two conditions

\(x \geq -9\) and \(x > 2\)

Take the stricter: \(x > 2\)

Domain: \(x > 2\)

Question 37

2.38 pts

Find the domain of the function:

\(f(x) = \frac{\sqrt{x+8}}{\sqrt{x-1}}\)

\(x > 1\)

\(x \geq 1\)

\(x > (-8)\)

\(x \neq 1\)

Explanation:

Solution — Domain:

The function: \(f(x) = \frac{\sqrt{x+8}}{\sqrt{x-1}}\)

📌 Step 1: Square root in numerator — the expression must be non-negative

\(x - -8 \geq 0 \;\Rightarrow\; x \geq -8\)

📌 Step 2: Square root in denominator — the expression must be strictly positive

\(x - 1 > 0 \;\Rightarrow\; x > 1\)

📌 Step 3: Intersection of the two conditions

\(x \geq -8\) and \(x > 1\)

Take the stricter: \(x > 1\)

Domain: \(x > 1\)

Question 38

2.38 pts

Find the domain of the function:

\(f(x) = \frac{\sqrt{x+7}}{\sqrt{x+3}}\)

\(x > (-3)\)

\(x \geq (-3)\)

\(x > (-7)\)

\(x \neq (-3)\)

Explanation:

Solution — Domain:

The function: \(f(x) = \frac{\sqrt{x+7}}{\sqrt{x+3}}\)

📌 Step 1: Square root in numerator — the expression must be non-negative

\(x - -7 \geq 0 \;\Rightarrow\; x \geq -7\)

📌 Step 2: Square root in denominator — the expression must be strictly positive

\(x - -3 > 0 \;\Rightarrow\; x > -3\)

📌 Step 3: Intersection of the two conditions

\(x \geq -7\) and \(x > -3\)

Take the stricter: \(x > -3\)

Domain: \(x > (-3)\)

Question 39

2.38 pts

Find the domain of the function:

\(f(x) = \frac{\sqrt{x-3}}{\sqrt{x+6}}\)

\(x \geq 3\)

\(x > 3\)

\(x > (-6)\)

\(x \neq (-6)\)

Explanation:

Solution — Domain:

The function: \(f(x) = \frac{\sqrt{x-3}}{\sqrt{x+6}}\)

📌 Step 1: Square root in numerator — the expression must be non-negative

\(x - 3 \geq 0 \;\Rightarrow\; x \geq 3\)

📌 Step 2: Square root in denominator — the expression must be strictly positive

\(x - -6 > 0 \;\Rightarrow\; x > -6\)

📌 Step 3: Intersection of the two conditions

\(x \geq 3\) and \(x > -6\)

Take the stricter: \(x \geq 3\)

Domain: \(x \geq 3\)

Question 40

2.38 pts

Find the domain of the function:

\(f(x) = \frac{\sqrt{x+2}}{\sqrt{x+8}}\)

\(x \geq (-2)\)

\(x > (-2)\)

\(x > (-8)\)

\(x \neq (-8)\)

Explanation:

Solution — Domain:

The function: \(f(x) = \frac{\sqrt{x+2}}{\sqrt{x+8}}\)

📌 Step 1: Square root in numerator — the expression must be non-negative

\(x - -2 \geq 0 \;\Rightarrow\; x \geq -2\)

📌 Step 2: Square root in denominator — the expression must be strictly positive

\(x - -8 > 0 \;\Rightarrow\; x > -8\)

📌 Step 3: Intersection of the two conditions

\(x \geq -2\) and \(x > -8\)

Take the stricter: \(x \geq -2\)

Domain: \(x \geq (-2)\)

Question 41

2.38 pts

Find the domain of the function:

\(f(x) = \frac{\sqrt{x+1}}{\sqrt{x+9}}\)

\(x \geq (-1)\)

\(x > (-1)\)

\(x > (-9)\)

\(x \neq (-9)\)

Explanation:

Solution — Domain:

The function: \(f(x) = \frac{\sqrt{x+1}}{\sqrt{x+9}}\)

📌 Step 1: Square root in numerator — the expression must be non-negative

\(x - -1 \geq 0 \;\Rightarrow\; x \geq -1\)

📌 Step 2: Square root in denominator — the expression must be strictly positive

\(x - -9 > 0 \;\Rightarrow\; x > -9\)

📌 Step 3: Intersection of the two conditions

\(x \geq -1\) and \(x > -9\)

Take the stricter: \(x \geq -1\)

Domain: \(x \geq (-1)\)

Question 42

2.38 pts

Find the domain of the function:

\(f(x) = \frac{\sqrt{x+5}}{\sqrt{x+6}}\)

\(x \geq (-5)\)

\(x > (-5)\)

\(x > (-6)\)

\(x \neq (-6)\)

Explanation:

Solution — Domain:

The function: \(f(x) = \frac{\sqrt{x+5}}{\sqrt{x+6}}\)

📌 Step 1: Square root in numerator — the expression must be non-negative

\(x - -5 \geq 0 \;\Rightarrow\; x \geq -5\)

📌 Step 2: Square root in denominator — the expression must be strictly positive

\(x - -6 > 0 \;\Rightarrow\; x > -6\)

📌 Step 3: Intersection of the two conditions

\(x \geq -5\) and \(x > -6\)

Take the stricter: \(x \geq -5\)

Domain: \(x \geq (-5)\)