Real-Valued Functions, Their Domain and Range

Correct answer: A. \([-1,\infty)\)

Since \(|x-6|\geq 0\), we have \(2|x-6|-1\geq -1\). At \(x=6\), \(|x-6|=0\), so \(f(6)=-1\), which is the minimum value. As \(|x-6|\) increases, the function increases without an upper bound. Therefore, the range is \([-1,\infty)\). Option B ignores the downward shift by 1. Exam tip: In \(a|x-h|+k\), if \(a>0\), the minimum value is \(k\).

Correct answer: A. ((-\infty,9])

Since (|x+1|\ge0), (9-3|x+1|\le9) and can go down without bound. In exams a negative coefficient changes the range direction.

Correct answer: A. ([-7,\infty))

(\sqrt{x+12}\ge0), so (f(x)\ge-7). In exams an outside subtraction shifts the square-root range downward.

Correct answer: A. ((-\infty,4])

(\sqrt{x-5}\ge0), so (4-\sqrt{x-5}\le4). In exams if a minus sign is before the square root, outputs go downward.

Correct answer: A. ([9,49])

On the negative interval, the smallest value of (x^2) is (9) and the largest is (49). In exams check endpoints when (0) is not in the interval.

Correct answer: A. ([0,36))

The domain contains (0), so the minimum is (0), and (6) is not included, so (36) is not included. In exams check how an open endpoint affects the range.

Correct answer: A. ((-13,3])

The function is decreasing; at (x=1) it gives (3), and as (x\to5) it approaches (-13), not included. In exams the range order reverses for a decreasing function.

Correct answer: A. ([0,7])

The domain contains (x=5), so the minimum is (0), and the far endpoint (12) gives maximum (7). In exams check the zero point of modulus.

Correct answer: A. ([1,5))

In this interval, (x+3) is negative, so (|x+3|=-(x+3)). (x=-4) gives included (1) and (x=-8) gives open (5).

Correct answer: A. ([0,4])

(\sqrt{x+4}) is an increasing function and gives (0) and (4) at the endpoints. In exams include endpoint values for a closed domain.

Correct answer: A. ([\frac{1}{8},\frac{1}{2}])

On this domain the denominator is positive and the reciprocal decreases. (x=3) gives (\frac{1}{2}) and (x=9) gives (\frac{1}{8}).

Correct answer: A. ([-1,-\frac{1}{4}])

The value of (x+2) lies in ([-4,-1]), so the reciprocal gives ([-1,-\frac{1}{4}]). In exams handle the order carefully for negative denominators.

Correct answer: A. (\mathbb{R}\setminus{4})

(\frac{1}{x-9}) is never (0), so output (4) cannot occur. In exams identify the excluded range value after a vertical shift.

Correct answer: A. (\mathbb{R}\setminus{-2})

(\frac{6}{x+5}) is never (0), so output (-2) is not obtained. In exams identify horizontal and vertical shifts of reciprocal functions separately.

Correct answer: A. (f(x)=\frac{x}{x^2-9})

In (\frac{x}{x^2-9}), the denominator is ((x-3)(x+3)), so (x=\pm3) are removed. In exams remove denominator zero values in rational functions.

Correct answer: A. (f(x)=\frac{2}{x-1}+5)

(\frac{2}{x-1}) is never (0), so output (5) cannot occur. In exams identify the range of a vertically shifted reciprocal.

Correct answer: A. Domain (\mathbb{R}) and range ([5,\infty))

Since (x^2+25\ge25), (f(x)\ge5) and every real (x) is allowed. In exams first check the minimum value inside.

Correct answer: A. Domain is (\mathbb{R}\setminus{-4,4})

The denominator (x^2-16) becomes (0) when (x=-4) or (x=4). In exams remove values that make the denominator (0).

Correct answer: A. (10)

(\sqrt{x+6}=4) gives (x+6=16), so (x=10). In exams check the answer in the original condition after squaring.

Correct answer: A. x=18 or x=-4

The governing concept is solving an absolute-value equation. Since f(x)=11, substitute this value to obtain |x-7|=11. An absolute value equal to a positive number has two branches: x-7=11 or x-7=-11. The first equation gives x=18, while the second gives x=-4. Both values are valid because |18-7|=11 and |-4-7|=11. Option A is therefore correct. Option D incorrectly keeps only the positive branch, and options B and C result from changing signs or subtracting incorrectly. Always consider both directions from 7 on the number line when solving |x-a|=b.

Correct answer: A. 8

Substituting \(x=-1\), \(f(-1)=(-1)^2-5(-1)+2=1+5+2=8\). Therefore, 8 is correct. The distractor \(-2\) may result from handling the sign of \(-5(-1)\) incorrectly. In exams, always use brackets when substituting a negative value.

Correct answer: A. \(-\frac{1}{4}\)

Substituting \(x=0\) gives \(f(0)=\frac{2(0)+1}{0-4}=\frac{1}{-4}=-\frac{1}{4}\). Option B results from overlooking the negative denominator \((-4)\). Exam tip: after substitution in a rational function, check the signs of both the numerator and denominator carefully.

Correct answer: A. (8)

(f(-10)=\sqrt{100-36}=\sqrt{64}=8). In exams the principal value of the square root is non-negative.

Correct answer: A. ([-15,\infty))

Since (x^2\ge0), (x^2-15\ge-15). In exams remember that the minimum value of the square function is (0).

Correct answer: A. Range is ([11,\infty)) and codomain is (\mathbb{R})

In the notation, the second set is the codomain (\mathbb{R}), while actual outputs are ([11,\infty)). In exams keep range and codomain separate.