Domain & Range Practice Quiz

The domain represents all possible input values (x-values) for which the function is defined. Understanding the domain is essential for determining which inputs can be used without causing undefined expressions.

The range is the set of all possible output values (y-values) that the function can produce. It is determined by the function's rule in combination with its domain.

The square root function √x is defined only for non-negative values of x. Therefore, the domain for f(x) = √x is all x such that x ≥ 0.

Since the square root function produces only non-negative outputs, the range of f(x) = √x is y ≥ 0. This means the output can be 0 or any positive number.

The function f(x) = 1/x is undefined when x equals 0 because division by zero is not allowed. Therefore, the domain consists of all real numbers except 0.

The function f(x) = (2x + 3)/(x - 4) is undefined when the denominator is zero, which occurs at x = 4. Therefore, the correct domain excludes x = 4.

The quadratic function f(x) = x² - 5 reaches its minimum value at x = 0, which gives f(0) = -5. Since x² is always non-negative, the range starts at -5 and continues to infinity.

For the square root √(x + 2) to be defined, the expression inside must be non-negative. Hence, x + 2 ≥ 0, which means x must be greater than or equal to -2.

The denominator x² - 9 factors to (x - 3)(x + 3) and equals zero when x = 3 or x = -3. These values must be excluded from the domain to avoid division by zero.

The absolute value function returns the non-negative magnitude of any input. Thus, the range of f(x) = |x| consists of all y such that y is greater than or equal to 0.

For the square root √(4 - x²) to yield a real number, the expression inside must be non-negative. Solving 4 - x² ≥ 0 gives the interval -2 ≤ x ≤ 2.

Since squaring any real number results in a non-negative value, the range of f(x) = x² is all y such that y ≥ 0.

For the logarithm to be defined, its argument must be positive. For f(x) = log(x - 1), the condition x - 1 > 0 leads to x > 1.

The function log(x) is continuous and unbounded as x varies over (0, ∞), meaning it can produce any real number. Therefore, its range is all real numbers.

The square root √(x - 3) requires that x - 3 ≥ 0, so x must be at least 3. Although the denominator x + 2 cannot be zero, x = -2 is already excluded by x ≥ 3, making the domain x ≥ 3.

For f(x) = √(4 - x²) the expression under the square root must be non-negative, which restricts x to [-2, 2]. The maximum value, 2, occurs when x = 0, and the minimum value 0 occurs at x = -2 and x = 2, setting the range as [0, 2].

The composite function (f ∘ g)(x) is f(√x) = 1/(x - 4), which requires that √x is defined (x ≥ 0) and that the denominator x - 4 ≠0, so x cannot equal 4. The domain is therefore all x ≥ 0 except x = 4.

For f(x) = log(5 - √x), the square root √x is defined for x ≥ 0 and the argument of the logarithm must be positive. Setting 5 - √x > 0 yields √x < 5, or x < 25. Combining these gives 0 ≤ x < 25.

The expression x² - 9 factors as (x - 3)(x + 3), which is zero when x equals 3 or -3. These values are excluded from the domain to prevent division by zero.

For f(x) = √(x + 1)/(x - 2), the square root requires x + 1 ≥ 0, so x must be at least -1, and the denominator x - 2 cannot be zero, eliminating x = 2. Analysis of the function on the intervals x ∈ [-1, 2) and x ∈ (2, ∞) shows that f(x) can produce any real number, making the range all real numbers.