Domain 4: Lesson 1 Fill-in-the-Blanks Quiz

A linear function like f(x) = 2x + 5 is defined for every real number since there are no restrictions on x. Therefore, the correct domain is all real numbers.

The square root function is only defined when its argument is non-negative. Thus, the domain of g(x) = √(x) requires x to be greater than or equal to 0.

Since division by zero is undefined, the function h(x) = 1/x cannot have x equal to 0. Hence, the domain includes all real numbers except 0.

A constant function does not impose any restrictions on the input variable x. Thus, the domain of f(x) = 7 is all real numbers.

The absolute value function |x| is defined for every real number. Therefore, the domain is all real numbers.

The expression under the square root must be non-negative, so x - 3 ≥ 0. Solving this inequality gives x ≥ 3, which is the correct domain.

The function f(x) = 1/(x - 2) is undefined when the denominator equals zero. Thus, x cannot be 2, and the domain includes all other real numbers.

For the square root to be defined, the radicand must be non-negative. Setting 4 - x ≥ 0 results in x ≤ 4, which is the correct domain.

The denominator factors as (x - 1)(x + 1), so it is zero when x is 1 or -1. Even though the numerator shares a factor with the denominator, these x values must be excluded from the domain.

The numerator √(x + 2) requires that x + 2 be non-negative, so x must be at least -2. Additionally, the denominator cannot be zero, which means x cannot equal 5. Combining these conditions gives the correct domain.

The square root in the denominator requires that its argument be positive to avoid division by zero. Thus, x - 1 must be greater than zero, which simplifies to x > 1.

The square root requires x to be non-negative while the denominator cannot be zero. Therefore, x must be greater than or equal to 0 and x cannot equal 3.

For the square root to be defined, the radicand 9 - 4x must be non-negative, which gives x ≤ 9/4. Additionally, x cannot equal -2 due to the denominator. The combination of these restrictions yields the correct domain.

Even though the expression can be simplified by canceling a common factor, the function is still undefined when x = 2 because it makes the denominator zero. Therefore, the correct domain excludes x = 2.

The expression under the square root must be non-negative, which leads to the inequality x² - 9 ≥ 0. Solving this gives two intervals, x ≤ -3 or x ≥ 3, which is the correct domain.

To determine the domain, analyze the inequality (x - 1)/(x + 3) ≥ 0. The critical points are x = -3 and x = 1, and a sign analysis shows that the expression is non-negative when x < -3 or x ≥ 1. Note that x = -3 is excluded because it would make the denominator zero.

The numerator √(x - 2) requires that x be at least 2, while the denominator √(5 - x) requires that 5 - x be positive, so x must be less than 5 to avoid a zero denominator. Combining these restrictions gives the domain 2 ≤ x < 5.

The first square root requires (x+4)/(x-2) to be non-negative, which is satisfied when x ≤ -4 or x > 2 (with x = 2 excluded). The second square root further restricts x by requiring 6 - x ≥ 0, meaning x ≤ 6. Combining these conditions yields x ≤ -4 or (2 < x ≤ 6).

Since the denominator contains √(x² - 16), the expression under the square root must be strictly positive to avoid division by zero, which requires x² - 16 > 0. Solving this inequality gives x < -4 or x > 4, which is the correct domain.

For the natural logarithm ln(x - 1) to be defined, its argument must be positive, so x > 1. Additionally, the square root √(10 - x) requires that 10 - x be non-negative, leading to x ≤ 10. The combination of these conditions results in the domain 1 < x ≤ 10.