Real-Valued Functions, Their Domain and Range

The direct answer is A, \\({0,3,8}\\). The domain is the finite set \\(A=\\{-4,-2,1\\}\\), so calculate the function for each input. For x=-4, \\(f(-4)=(-4)^2+2(-4)=16-8=8\\). For x=-2, \\(f(-2)=4-4=0\\). For x=1, \\(f(1)=1+2=3\\). The range is the set of distinct outputs, \\({8,0,3}={0,3,8}\\). Option A is correct. Option B merely repeats the domain, not the outputs. Option C has -4, which is an input, not an output; the correct output for -2 is 0, not -4. Option D gives values that do not result from the formula. Memory cue: for a finite domain, substitute every listed input and collect the images.

For the given inputs, the outputs are (0,1,2,3). In exams use only the given elements in a set domain.

The outputs are (5,2,0,3), written as ({0,2,3,5}) in set form. In exams write repeated or unordered values in set form.

For a real-valued square-root function, the radicand must be non-negative, and its denominator must be nonzero. Thus we require (6−x)/(x+2)≥0 with x≠−2. The critical values are x=−2, where the expression is undefined, and x=6, where the numerator is zero. A sign check gives: for x<−2, the numerator is positive and denominator negative, so the quotient is negative; for −2<x<6, both are positive, so the quotient is positive; for x>6, the quotient is negative. At x=6 the radicand is zero and is allowed. Therefore the domain is (−2,6], so option A is correct. D wrongly includes −2, while B and C select intervals where the radicand is negative.

The expression under the square root must be at least zero, and the denominator must not vanish. Hence we solve (x+5)/(9−x)≥0 with x≠9. The critical numbers are −5, where the numerator is zero, and 9, where the denominator is zero. On x<−5, the numerator is negative and denominator positive, giving a negative quotient. On −5<x<9, both numerator and denominator are positive, giving a non-negative quotient. For x>9, the numerator is positive but denominator negative, again giving a negative quotient. The value −5 is included because the radicand is zero there; 9 is excluded because division by zero is undefined. Thus the domain is [−5,9), option A. C incorrectly includes 9, B reverses the valid interval, and D ignores the square-root condition.

The numerator √(x−2) is real only when x−2≥0, which gives x≥2. The denominator √(8−x) must satisfy two conditions: its radicand must be non-negative and the denominator itself must not be zero. Because it is in the denominator, we need √(8−x)>0, equivalently 8−x>0, so x<8. Combining x≥2 with x<8 gives 2≤x<8, or [2,8). The endpoint x=2 is allowed because the numerator is zero while the denominator is √6, which is nonzero. The endpoint x=8 is forbidden because the denominator becomes zero. Therefore option A is correct. B unnecessarily removes 2, C includes an undefined value, and D reverses the relevant inequalities.

Both square roots impose conditions. The numerator \(\sqrt{11-x}\u0005 is real when \(11-x\ge0\u0005, giving \(x\le11\u0005. The denominator \(\sqrt{x+1}\u0005 must be real and nonzero. Thus \(x+1>0\u0005, giving \(x>-1\u0005. These conditions must be satisfied simultaneously.

Combining them gives \(-1<x\le11\u0005, or \((-1,11]\u0005. Therefore option A is correct. The value \(x=-1\u0005 is excluded because the denominator becomes zero. The value \(x=11\u0005 is included because the numerator may equal zero while the denominator remains nonzero.

Since (x=2) is included, output (0) is included, but (x=18) is open, so (4) is not included. In exams track endpoint openness in the range.

Since (x=20) is included, output (0) is included, and since (x=4) is open, output (4) is not included. In exams endpoints reverse for a decreasing square-root expression.

For (\sqrt{9-x^2}) to be real, (9-x^2\ge 0), so (x\in[-3,3]) and (f(x)\in[0,3]). In exams, always keep the expression inside a square root (\ge 0).

Complete the square to expose the function’s minimum: f(x)=x²−4x+7=(x−2)²+3. For every real x, the square (x−2)² is non-negative, so f(x) is always at least 3. When x=2, the square term is zero and f(2)=3, so the lower endpoint is attained and included. As x moves farther from 2 in either direction, (x−2)² can become arbitrarily large, so f(x) has no upper bound. Therefore the range is [3,∞), making option A correct. Option B reverses the direction of the bound and describes the wrong type of parabola. Option C incorrectly treats the constant term 7 as the minimum, and option D ignores the definite lower bound produced by the squared term. The vertex form gives both the minimum value and the endpoint inclusion immediately.

Rewrite the quadratic in vertex form: f(x)=−x²+6x−5=−(x²−6x+9)+4=−(x−3)²+4. Since (x−3)² is always non-negative, its negative is always at most zero. Thus f(x)≤4 for every real x. Equality occurs at x=3, where the square term is zero and f(3)=4, so 4 is included in the range. As |x| becomes arbitrarily large, the negative square term decreases without bound, meaning the function can take values as small as desired. Therefore the range is (−∞,4], which is option A. Option B has the direction suitable for an upward-opening parabola; C confuses the constant term with the maximum, and D incorrectly claims that every real value occurs.

The governing concept is the range of a square-root function. A square root always produces a non-negative output, so f(x) cannot be less than 0. Also, the expression inside the root is 9 − (x − 1)². Its largest possible value is 9, attained when x = 1. Therefore the largest function value is √9 = 3. The value 0 is also attained when (x − 1)² = 9, namely at x = −2 or x = 4. As x varies over the domain [−2, 4], the inside expression covers every value from 0 to 9, so the output covers every value from 0 to 3. Hence the range is [0, 3], making option A correct. Option B incorrectly includes negative outputs, option C excludes the attained value 0, and option D confuses the radicand with the function value.

The governing concept is finding a domain by applying every restriction in the formula. For log₃x to exist, its argument must be positive, so x > 0. In addition, the logarithm is in the denominator, and a denominator cannot be zero. Therefore log₃x ≠ 0. Since log₃1 = 0, we must exclude x = 1. Combining both conditions gives x > 0 and x ≠ 1, or the interval set (0, ∞) − {1}. Notice that all positive numbers other than 1 are allowed; for example, log₃3 = 1 and log₃(1/3) = −1 are both defined and nonzero. Thus option A is correct. Option B incorrectly includes 1, option C incorrectly includes non-positive numbers, and option D wrongly excludes values between 0 and 1.

The domain must satisfy the restrictions from both terms simultaneously. For √(x − 1) to be real, the radicand must be non-negative: x − 1 ≥ 0, so x ≥ 1. The square-root endpoint x = 1 is allowed because √0 exists. For the rational term 1/(x − 5), the denominator must not vanish, so x ≠ 5. Intersecting these conditions gives [1, ∞) with the single point 5 removed, written [1, ∞) − {5}. Therefore option A is correct. Option B wrongly excludes the valid endpoint 1; option C excludes all values at least 5 even though they are valid except for 5; and option D includes values below 1, where the square root is not real. Both restrictions are essential.

Complete the square in the denominator: x^2 + 2x + 2 = (x + 1)^2 + 1. Since (x + 1)^2 is always non-negative, the denominator is at least 1, with equality when x = −1. The denominator is therefore positive and has its minimum value 1. For a positive reciprocal, the largest value occurs when the denominator is smallest. Consequently f(−1) = 1/1 = 1, so the maximum value is 1. Option A is correct. The value 1/2 would arise only if the denominator had minimum 2, which it does not; 2 is not the maximum, and 0 is approached only as the denominator grows without bound, not attained as the maximum.

For a real-valued function, the expression under a square root must be non-negative. Here the square root is also in the denominator, so its value cannot be zero. Consequently, the stricter condition is x² − 9 > 0, not merely x² − 9 ≥ 0. Factoring gives (x − 3)(x + 3) > 0. A sign analysis shows this inequality holds when x < −3 or x > 3. Therefore the domain is (−∞, −3) ∪ (3, ∞), which is option A. The endpoints −3 and 3 must be excluded because they make the denominator zero. Option B incorrectly includes those endpoints, option C makes the radicand non-positive, and option D excludes only the endpoints while admitting values between them.

First, the radicand must satisfy 4 − x² ≥ 0, so the domain is −2 ≤ x ≤ 2. Since f(x) is a principal square root, its output can never be negative. The largest value of 4 − x² occurs when x = 0, giving f(0) = √4 = 2. At x = −2 or x = 2, the value is √0 = 0. As x varies through the domain, the function takes every value from 0 to 2; geometrically, its graph is the upper semicircle x² + y² = 4. Therefore the range is [0, 2], so option A is correct. Option B wrongly includes negative outputs, option C confuses the radicand with the function value, and option D includes values below zero.

The governing idea is to rewrite the quadratic denominator by completing the square: x² + 4x + 5 = (x + 2)² + 1. Since (x + 2)² is always non-negative, the denominator is at least 1 and is exactly 1 when x = −2. Therefore f(x) is positive, never exceeds 1, and equals 1 at x = −2. As |x| becomes very large, the denominator increases without bound, so f(x) approaches 0 but never equals 0. Every value strictly between 0 and 1 is obtained because the denominator can take every value from 1 upward. Hence the range is (0, 1]. Option B incorrectly gives the denominator-type interval, while C includes impossible negative values and D wrongly includes values greater than 1.

The key point is that cancellation does not change the original domain restriction. Factor the numerator: x² − 4 = (x − 2)(x + 2). Since the denominator requires x ≠ 2, the function simplifies to f(x) = x + 2 only for x ≠ 2. The linear expression x + 2 can produce every real value except 4: obtaining output 4 would require x + 2 = 4, hence x = 2, but that input is excluded from the original function. Every other real output y is achieved by choosing x = y − 2, which is allowed whenever y ≠ 4. Thus the range is ℝ \ {4}. Option B confuses the excluded input with the excluded output; option C forgets the hole, and option D incorrectly treats the function as bounded.

The domain must satisfy two independent conditions. First, the square root √x is real only when x ≥ 0. Second, the denominator cannot be zero, so 2 − √x ≠ 0. Solving the second condition gives √x ≠ 2, and therefore x ≠ 4. Combining these restrictions produces [0, ∞) with the single point 4 removed. Notice that x = 0 is allowed: the denominator is then 2, not zero. Values greater than 4 are also allowed, although the denominator becomes negative; a negative denominator is perfectly valid. Thus option A is correct. Option B wrongly excludes all x ≥ 4, option C incorrectly removes 0, and option D includes negative numbers for which the square root is not real.

For every real x, x² ≥ 0. If a > 0, then ax² ≥ 0, so f(x) = ax² + 4 is at least 4. The minimum value 4 is attained at x = 0. Moreover, as |x| increases, ax² can take every non-negative value, so the range is exactly [4, ∞). If a < 0, the parabola opens downward and its range is (−∞, 4], not the stated interval. If a = 0, the function is constant and its range is only {4}, not an unbounded interval. Therefore the necessary and sufficient condition is a > 0. Option D is too broad because it also includes negative a, which produces the opposite type of range. The conclusion follows from the sign of the quadratic coefficient.

A logarithm log₂x is defined only when its argument is positive, so x > 0. There is an additional restriction because the logarithm appears in the denominator: log₂x must not equal zero. Since log₂1 = 0, x = 1 must be excluded. No other positive x is excluded, because the base 2 is valid and log₂x is defined for every x > 0. Consequently, the domain is (0, ∞) \ {1}. Option B includes x = 1 even though division by zero would occur. Option C wrongly allows negative x, where the real logarithm is undefined, and option D omits valid values between 0 and 1. Both the logarithm condition and denominator condition are essential.

Since x² ≥ 0 for every real x, we have x² + 1 ≥ 1. Taking square roots gives √(x² + 1) ≥ 1. Therefore 5 − √(x² + 1) ≤ 4, and equality occurs at x = 0, so the upper endpoint 4 is included. As |x| tends to infinity, √(x² + 1) becomes arbitrarily large, making f(x) decrease without bound. Thus all values less than or equal to 4 occur, and the range is (−∞, 4]. Option B reverses the direction of the inequality. Option C incorrectly includes values between 4 and 5, although the square-root term is never below 1. Option D imposes an unjustified lower bound of 0; the function can be negative and can become arbitrarily small.

The denominator |x| must be nonzero, so x = 0 is excluded from the domain. For every positive x, |x| = x and therefore x/|x| = 1. For every negative x, |x| = −x and therefore x/|x| = −1. Thus every allowed input produces one of the two values −1 or 1, and both values actually occur because positive and negative real inputs exist. Hence the domain is R \ {0} and the range is {−1, 1}; option A is correct. Option B incorrectly includes zero, where the expression is undefined. Option C gives an interval instead of the two attained values, and option D wrongly claims that all real outputs are possible.