Function Domain — Dynamic Practice
Function Domain — Dynamic Practice. Practice questions to deepen understanding of the domain of functions. Online dynamic learning math system.
Dynamic practice in finding the domain of functions — roots, denominators, intervals, and algebraic restrictions. New questions every attempt.
Question 1
7.69 pts
📐 Domain:
Find the domain of the function:
\(f(x) = \sqrt-x + 6\)
Find the domain of the function:
\(f(x) = \sqrt-x + 6\)
Explanation:
Solution:
A square root is defined when the expression under it is non-negative:
\(-x + 6 \geq 0\)
\(-1x \geq -6\)
\(x \leq 6\)
Note: when dividing by a negative number, the sign flips!
Domain on the number line: Answer: \(x \leq 6\)
A square root is defined when the expression under it is non-negative:
\(-x + 6 \geq 0\)
\(-1x \geq -6\)
\(x \leq 6\)
Note: when dividing by a negative number, the sign flips!
Domain on the number line: Answer: \(x \leq 6\)
Question 2
7.69 pts
📐 Domain:
Find the domain of the function:
\(f(x) = \sqrt-x + 0\)
Find the domain of the function:
\(f(x) = \sqrt-x + 0\)
Explanation:
Solution:
A square root is defined when the expression under it is non-negative:
\(-x + 0 \geq 0\)
\(-1x \geq 0\)
\(x \leq 0\)
Note: when dividing by a negative number, the sign flips!
Domain on the number line: Answer: \(x \leq 0\)
A square root is defined when the expression under it is non-negative:
\(-x + 0 \geq 0\)
\(-1x \geq 0\)
\(x \leq 0\)
Note: when dividing by a negative number, the sign flips!
Domain on the number line: Answer: \(x \leq 0\)
Question 3
7.69 pts
📐 Domain:
Find the domain of the function:
\(f(x) = \sqrtx - 8\)
Find the domain of the function:
\(f(x) = \sqrtx - 8\)
Explanation:
Solution:
A square root is defined when the expression under it is non-negative:
\(x - 8 \geq 0\)
\(1x \geq 8\)
\(x \geq 8\)
Domain on the number line: Answer: \(x \geq 8\)
A square root is defined when the expression under it is non-negative:
\(x - 8 \geq 0\)
\(1x \geq 8\)
\(x \geq 8\)
Domain on the number line: Answer: \(x \geq 8\)
Question 4
7.69 pts
📐 Domain – Parabola:
Find the domain of the function:
\(f(x) = \sqrtx^2 + 2x - 3\)
Find the domain of the function:
\(f(x) = \sqrtx^2 + 2x - 3\)
Explanation:
Solution:
A square root is defined when the expression under it is non-negative:
\(x^2 + 2x - 3 \geq 0\)
Step 1: Calculate the discriminant:
\(\Delta = 2^2 - 4 \cdot (1) \cdot (-3) = 16\)
Step 2: Find the roots: \(x_1 = -3, x_2 = 1\)
Step 3: The parabola opens upward (a > 0) → the expression is positive outside the interval between the roots
Diagram: Answer: \(x \leq -3\) or \(x \geq 1\)
A square root is defined when the expression under it is non-negative:
\(x^2 + 2x - 3 \geq 0\)
Step 1: Calculate the discriminant:
\(\Delta = 2^2 - 4 \cdot (1) \cdot (-3) = 16\)
Step 2: Find the roots: \(x_1 = -3, x_2 = 1\)
Step 3: The parabola opens upward (a > 0) → the expression is positive outside the interval between the roots
Diagram: Answer: \(x \leq -3\) or \(x \geq 1\)
Question 5
7.69 pts
📐 Domain – Parabola:
Find the domain of the function:
\(f(x) = \sqrtx^2 + 3x - 4\)
Find the domain of the function:
\(f(x) = \sqrtx^2 + 3x - 4\)
Explanation:
Solution:
A square root is defined when the expression under it is non-negative:
\(x^2 + 3x - 4 \geq 0\)
Step 1: Calculate the discriminant:
\(\Delta = 3^2 - 4 \cdot (1) \cdot (-4) = 25\)
Step 2: Find the roots: \(x_1 = -4, x_2 = 1\)
Step 3: The parabola opens upward (a > 0) → the expression is positive outside the interval between the roots
Diagram: Answer: \(x \leq -4\) or \(x \geq 1\)
A square root is defined when the expression under it is non-negative:
\(x^2 + 3x - 4 \geq 0\)
Step 1: Calculate the discriminant:
\(\Delta = 3^2 - 4 \cdot (1) \cdot (-4) = 25\)
Step 2: Find the roots: \(x_1 = -4, x_2 = 1\)
Step 3: The parabola opens upward (a > 0) → the expression is positive outside the interval between the roots
Diagram: Answer: \(x \leq -4\) or \(x \geq 1\)
Question 6
7.69 pts
Find the domain of the function:
\(f(x) = \frac{x}{x-10}\)
\(f(x) = \frac{x}{x-10}\)
Explanation:
Solution — Domain:
The function: \(f(x) = \frac{x}{x-10}\)
The function: \(f(x) = \frac{x}{x-10}\)
📌 Step 1: The denominator cannot be zero
\(x - 10 \neq 0\)
📌 Step 2: Isolate x
\(x \neq 10\)
Domain: \(x \neq 10\)
Question 7
7.69 pts
Find the domain of the function:
\(f(x) = \frac{1}{x+6}\)
\(f(x) = \frac{1}{x+6}\)
Explanation:
Solution — Domain:
The function: \(f(x) = \frac{1}{x+6}\)
The function: \(f(x) = \frac{1}{x+6}\)
📌 Step 1: The denominator cannot be zero
\(x + 6 \neq 0\)
📌 Step 2: Isolate x
\(x \neq -6\)
Domain: \(x \neq -6\)
Question 8
7.69 pts
Find the domain of the function:
\(f(x) = \frac{\sqrt{x+4}}{\sqrt{x-1}}\)
\(f(x) = \frac{\sqrt{x+4}}{\sqrt{x-1}}\)
Explanation:
Solution — Domain:
The function: \(f(x) = \frac{\sqrt{x+4}}{\sqrt{x-1}}\)
The function: \(f(x) = \frac{\sqrt{x+4}}{\sqrt{x-1}}\)
📌 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
\(x - 1 > 0 \;\Rightarrow\; x > 1\)
📌 Step 3: Intersection of the two conditions
\(x \geq -4\) and \(x > 1\)
Take the stricter: \(x > 1\)
Domain: \(x > 1\)
Question 9
7.69 pts
Find the domain of the function:
\(f(x) = \sqrt{\frac{x+3}{x+1}}\)
\(f(x) = \sqrt{\frac{x+3}{x+1}}\)
Explanation:
Solution — Domain:
The function: \(f(x) = \sqrt{\frac{x+3}{x+1}}\)
The function: \(f(x) = \sqrt{\frac{x+3}{x+1}}\)
📌 Step 1: The expression under the root must be non-negative
\(\dfrac{x - -3}{x - -1} \geq 0\)
(The denominator must not be zero: \(x \neq -1\))
📌 Step 2: Roots — x=-3 and x=-1
The quotient is non-negative when the numerator and denominator have the same sign
📌 Step 3: Sign table
| x | x < -3 | -3 | -3 < x < -1 | -1 | x > -1 |
|---|---|---|---|---|---|
| Quotient sign | + | 0 | − | ∄ | + |
Include \(x = -3\) (the quotient = 0, allowed), exclude \(x = -1\)
Domain: \(x \leq -3 \text{ or } x > -1\)
Question 10
7.69 pts
Find the domain of the function:
\(f(x) = \frac{1}{\sqrt{x+4}}\)
\(f(x) = \frac{1}{\sqrt{x+4}}\)
Explanation:
Solution — Domain:
The function: \(f(x) = \frac{1}{\sqrt{x+4}}\)
The function: \(f(x) = \frac{1}{\sqrt{x+4}}\)
📌 Step 1: Square root in denominator — the expression must be strictly positive
\(x + 4 > 0\)
📌 Step 2: Isolate x
\(x > -4\)
⚠️ Equality not included! Zero in the denominator = undefined
Domain: \(x > -4\)
Question 11
7.69 pts
Find the domain of the function:
\(f(x) = \frac{1}{\sqrt{(x-2)(x-6)}}\)
\(f(x) = \frac{1}{\sqrt{(x-2)(x-6)}}\)
Explanation:
Solution — Domain:
The function: \(f(x) = \frac{1}{\sqrt{(x-2)(x-6)}}\)
The function: \(f(x) = \frac{1}{\sqrt{(x-2)(x-6)}}\)
📌 Step 1: Square root in denominator — the product must be strictly positive
\((x - 2)(x - 6) > 0\)
📌 Step 2: Roots of the expression
\(x = 2\) or \(x = 6\)
📌 Step 3: Sign table (upward-opening parabola — positive outside the roots)
| x | x < 2 | 2 | 2 < x < 6 | 6 | x > 6 |
|---|---|---|---|---|---|
| Sign | + | 0 | − | 0 | + |
Strict positivity required: \(x < 2\) or \(x > 6\)
Domain: \(x < 2 \text{ or } x > 6\)
Question 12
7.69 pts
Find the domain of the function:
\(f(x) = \frac{\sqrt{x-2}}{\sqrt{x-3}}\)
\(f(x) = \frac{\sqrt{x-2}}{\sqrt{x-3}}\)
Explanation:
Solution — Domain:
The function: \(f(x) = \frac{\sqrt{x-2}}{\sqrt{x-3}}\)
The function: \(f(x) = \frac{\sqrt{x-2}}{\sqrt{x-3}}\)
📌 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
\(x - 3 > 0 \;\Rightarrow\; x > 3\)
📌 Step 3: Intersection of the two conditions
\(x \geq 2\) and \(x > 3\)
Take the stricter: \(x > 3\)
Domain: \(x > 3\)
Question 13
7.69 pts
Find the domain of the function:
\(f(x) = \frac{x}{x+10}\)
\(f(x) = \frac{x}{x+10}\)
Explanation:
Solution — Domain:
The function: \(f(x) = \frac{x}{x+10}\)
The function: \(f(x) = \frac{x}{x+10}\)
📌 Step 1: The denominator cannot be zero
\(x + 10 \neq 0\)
📌 Step 2: Isolate x
\(x \neq -10\)
Domain: \(x \neq -10\)