Real-Valued Functions, Their Domain and Range

Correct answer: A. Domain ([1,\infty)), range ([3,\infty))

The conditions (x-1\ge 0) and (\sqrt{x-1}\ge 0) give the answer. The inside restriction gives the domain, and the outside shift changes the range.

Correct answer: A. Domain ((-\infty,4]), range ([1,\infty))

From (4-x\ge 0), (x\le 4), and (\sqrt{4-x}\ge 0), the range is ([1,\infty)). Checking equality at endpoints is important.

Correct answer: A. (-2, ∞)

The governing rule is that a square root requires its radicand to be non-negative, while a denominator must be non-zero. Here, x + 2 occurs under a square root and the entire square root is in the denominator, so we need x + 2 > 0, not merely x + 2 ≥ 0. Solving gives x > -2. Therefore, the domain is (-2, ∞), including every real number greater than -2 but excluding -2 itself, because at x = -2 the denominator becomes zero. The added constant 5 changes output values only and does not impose any new restriction. Hence option A is correct; option B incorrectly includes -2, while C reverses the inequality and D excludes only one point without applying the radical condition.

Correct answer: A. ((0,1])

The denominator (x^2-6x+10=(x-3)^2+1) is at least (1). Therefore the reciprocal is greater than (0) and at most (1).

Correct answer: A. ((0,\frac{1}{4}])

The denominator has minimum value (4), so the maximum function value is (\frac{1}{4}). For large denominators the function approaches (0) but never becomes (0).

Correct answer: A. \\([-1,2]\\)

For real values, \(9-x^2\\ge 0\), so \(-3\\le x\le 3\). Hence \(\\sqrt{9-x^2}\) takes every value in \[0,3\]. Letting \(y=\\sqrt{9-x^2}\), we have \(f(x)=2-y\), so the range is \[2-3,2-0\\]=\[-1,2\]. Option B would result from adding rather than subtracting, and both endpoints must be included.

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

Here (f(x)=x+1), but (x=1) is not allowed, so (y=2) is not obtained. After simplification, always track excluded domain values.

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

Since \(x^2-4=(x-2)(x+2)\), for \(x\ne2\) we have \(f(x)=x+2\). The value \(4\) would occur only when \(x=2\), but that input is excluded from the domain. Conversely, for every \(y\ne4\), choosing \(x=y-2\ne2\) gives \(f(x)=y\). Hence the range is \(\mathbb{R}\setminus\{4\}\). Exam tip: after simplifying a rational function, check whether an excluded input removes a corresponding output value.

Correct answer: A. ([-3,5])

Both square roots require (x+3\ge 0) and (5-x\ge 0) simultaneously. Take the intersection of the conditions.

Correct answer: A. ((-1,4])

A real-valued function can use a square root only when its radicand is non-negative. However, a square root in a denominator must also be nonzero, because division by zero is not defined. Thus the first term requires \(x+1>0\), not merely \(x+1\ge0\). The second term requires \(4-x\ge0\). These conditions must hold together.

Solving the first inequality gives \(x>-1\), while the second gives \(x\le4\). Their intersection is \(-1<x\le4\), written as \((-1,4]\). The value \(-1\) is excluded because it makes the denominator zero, whereas \(4\) is included because the second square root is then zero and the first term remains defined. Therefore option A is correct.

Correct answer: A. [−3, 2)

The radicand of a real square root must satisfy (x + 3)/(2 − x) ≥ 0, and x = 2 must be excluded because it makes the denominator zero. The critical points are −3, where the numerator is zero, and 2, where the denominator is zero. A sign chart shows the quotient is negative for x < −3, non-negative on [−3, 2), and negative for x > 2. The point −3 is included because the quotient equals zero and its square root exists. The point 2 is excluded because the original fraction is undefined. Therefore the domain is [−3, 2), which is option A. Option D wrongly excludes −3, option B includes the forbidden denominator value 2, and option C selects intervals where the radicand is negative.

Correct answer: A. \(\mathbb{R}\setminus\{1,-2\}\)

In a rational function, every denominator must be non-zero. Here, \(x-1=0\) gives \(x=1\), and \(x+2=0\) gives \(x=-2\); hence both values must be excluded from the domain. Therefore, the domain is \(\mathbb{R}\setminus\{1,-2\}\). Option B incorrectly changes \(-2\) to \(2\). Exam tip: set each denominator equal to zero and exclude all resulting values from the real numbers.

Correct answer: B. \([1,5)\cup(5,\infty)\)

For \(f(x)=\dfrac{\sqrt{x-1}}{x-5}\), two conditions:
1) Square root: \(x-1\ge 0\Rightarrow x\ge 1\).
2) Denominator not zero: \(x-5\neq 0\Rightarrow x\neq 5\).
Combine: \(x\in[1,\infty)\) but excluding \(x=5\). Hence domain is \([1,5)\cup(5,\infty)\).

Correct answer: A. \(( -\infty,-4]\cup[-2,\infty)\)

For the square root to be real, the radicand must be non-negative. Since \(x^2+6x+8=(x+4)(x+2)\), we solve \((x+4)(x+2)\ge 0\). This gives \(x\le -4\) or \(x\ge -2\). Hence, the domain is \(( -\infty,-4]\cup[-2,\infty)\). Option B is incorrect because the product is negative between \(-4\) and \(-2\). Exam tip: For a radical function, require the expression inside the root to be at least zero and use a sign chart.

Correct answer: A. x = 2

For a rational function, the domain contains all real inputs for which the denominator is non-zero. Factor the denominator: x² - 5x + 6 = (x - 2)(x - 3). It becomes zero when x = 2 or x = 3, so both values must be excluded from the domain. Among the four choices, only x = 2 is an excluded value; x = 0 gives denominator 6, x = 1 gives denominator 2, and x = 4 gives denominator 2, so each of those produces a defined function value. Therefore option A is the correct response. Although x = 3 is also outside the domain, it is not listed, so there is no ambiguity in the given options.

Correct answer: A. (\frac{3}{4})

The denominator (x^2+2x+5=(x+1)^2+4) has minimum value (4). A positive fraction is maximum when its denominator is minimum.

Correct answer: A. \((-4,-2]\)

Since \((x+3)^2\ge 0\), we get \((x+3)^2+1\ge 1\). Hence, \(0<\frac{2}{(x+3)^2+1}\le 2\). At \(x=-3\), the fraction attains its maximum value \(2\), so \(f(x)\) attains the maximum value \(-2\). As \(|x|\to\infty\), the fraction approaches \(0\), so \(f(x)\) approaches \(-4\) but never equals it. Therefore, the range is \((-4,-2]\). Option B is incorrect because it includes \(-4\). Exam tip: For expressions involving a squared term in the denominator, first find the minimum possible denominator.

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

Piecewise checking gives (-4) for (x\le -2), (4) for (x\ge 2), and (2x) between them. Hence all values in ([-4,4]) occur.

Correct answer: A. (4)

The distance between the points (1) and (-3) is (4). The minimum sum of absolute distances often equals the distance between fixed points.

Correct answer: A. Domain (\mathbb{R}\setminus{0}), range ({-1,1})

The denominator (|x|) must be non-zero, so (x\ne 0). Positive (x) gives (1), and negative (x) gives (-1).

Correct answer: A. \([0,\infty)\)

Since \(x^2+4x+4=(x+2)^2\), we get \(f(x)=\sqrt{(x+2)^2}=|x+2|\). Its minimum value is \(0\), attained at \(x=-2\), and it can take every non-negative value thereafter. Therefore, the range is \([0,\infty)\). Option B is incorrect because it excludes \(0\). Exam tip: Rewrite the expression under the square root as a perfect square and identify its minimum value.

Correct answer: A. [√7, ∞)

The governing idea is that a square is always non-negative. Thus (x - 2)^2 ≥ 0 for every real x, so (x - 2)^2 + 7 ≥ 7. Taking the non-negative square root gives f(x) ≥ √7. Equality is actually attained when x - 2 = 0, that is, at x = 2; hence √7 must be included in the range. As x moves away from 2, the square term can become arbitrarily large, so the function values increase without bound. Consequently the range is [√7, ∞). Option B incorrectly uses 7 instead of its square root, option C wrongly excludes the attainable minimum, and option D ignores the lower bound. Therefore option A is correct.

Correct answer: A. ((0,\frac{1}{2}])

The denominator (|x+1|+2) has minimum (2) and can grow without bound. Thus the function is greater than (0) and at most (\frac{1}{2}).

Correct answer: A. ([0,1))

Let (t=|x|), so (t\ge 0) and (f=\frac{t}{t+1}). At (t=0) the value is (0), and as (t) grows it approaches (1) but never reaches it.

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

Here (f(x)=1-\frac{1}{x^2+2}) and (x^2+2\ge 2). Thus the minimum is (\frac{1}{2}) and (1) is not attained.