Problems on finding the domain of a function are presented with their solutions.
Solution to Problems
Problem 1
Find the domain of the function defined by:
\[ f(x) = x + 1 \]View Solution
This is a linear (polynomial) function. Polynomial functions are defined for all real numbers. Hence, its domain is:
\[ (-\infty, +\infty) \]Problem 2
Find the domain of the function defined by:
\[ f(x) = \sqrt{2x} \]View Solution
This is a square root function. The expression under the square root must be non-negative:
\[ 2x \geq 0 \implies x \geq 0 \]The solution set for the inequality gives the domain in interval form:
\[ [0, +\infty) \]Problem 3
Find the domain of the function defined by:
\[ f(x) = \frac{x - 1}{x - 3} \]View Solution
This is a rational function. Its domain is the set of all real numbers except those values of \( x \) that make the denominator zero:
\[ x - 3 = 0 \implies x = 3 \]Excluding \( x = 3 \), the domain is given by the interval:
\[ (-\infty, 3) \cup (3, +\infty) \]Problem 4
Find the domain of the function defined by:
\[ f(x) = \frac{\sqrt{-x + 1}}{x + 3} \]View Solution
To find the domain of this function, we must satisfy two conditions simultaneously:
- Condition 1: The expression under the square root must be non-negative: \[ -x + 1 \geq 0 \implies x \leq 1 \]
- Condition 2: The denominator must be non-zero: \[ x + 3 \neq 0 \implies x \neq -3 \]
Combining both conditions, the domain of the function is:
\[ (-\infty, -3) \cup (-3, 1] \]Problem 5
Find the domain of the function defined by:
\[ f(x) = \sqrt[3]{2x + 1} \]View Solution
This is a cube root (odd root) function. The expression \( 2x + 1 \) can take any real value because odd roots are defined for all real numbers. Hence, the domain is:
\[ (-\infty, +\infty) \]Problem 6
Find the domain of the function defined by:
\[ f(x) = \ln(x^2 - 9) \]View Solution
The argument of the logarithmic function must be strictly positive for the function to be real-valued:
\[ x^2 - 9 > 0 \]Factoring the left side:
\[ (x - 3)(x + 3) > 0 \]Solving this polynomial inequality yields the domain:
\[ (-\infty, -3) \cup (3, +\infty) \]Problem 7
Find the domain of the function defined by:
\[ f(x) = 2\sin(x - 1) \]View Solution
The sine function is defined for all real numbers. The argument \( x - 1 \) can be any real number, so the domain is:
\[ (-\infty, +\infty) \]Problem 8
Find the domain of the function defined by:
\[ f(x) = e^{x - 4} \]View Solution
The natural exponential function is defined for all real exponents. The exponent \( x - 4 \) can take any real value, so the domain is:
\[ (-\infty, +\infty) \]Problem 9
Find the domain of the function defined by:
\[ f(x) = \arcsin(x^2 - 1) \]View Solution
For the arcsine function to be real-valued, its input must lie within the closed interval \([-1, 1]\):
\[ -1 \leq x^2 - 1 \leq 1 \]Add 1 to all parts of the inequality:
\[ 0 \leq x^2 \leq 2 \]Taking the square root of all parts yields:
\[ -\sqrt{2} \leq x \leq \sqrt{2} \]Thus, the domain in interval form is:
\[ [-\sqrt{2}, \sqrt{2}] \]Problem 10
Find the domain of the function defined by:
\[ f(x) = \frac{1}{x^3 + x^2 - 2x} \]View Solution
The domain is restricted by excluding values that make the denominator equal to zero. Set the denominator to zero:
\[ x^3 + x^2 - 2x = 0 \]Factor out \( x \):
\[ x(x^2 + x - 2) = 0 \]Further factor the quadratic expression:
\[ x(x + 2)(x - 1) = 0 \]The roots are \( x = 0 \), \( x = -2 \), and \( x = 1 \). Excluding these values, the domain of the function is:
\[ (-\infty, -2) \cup (-2, 0) \cup (0, 1) \cup (1, +\infty) \]