Real-Valued Functions, Their Domain and Range

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

The minimum value of (|x+2|) is (0), so the output is at least (1). In exams the multiplier (3) does not change the minimum when modulus is (0).

Correct answer: A. ([-6,\infty))

(\sqrt{x+4}\ge 0), so (\sqrt{x+4}-6\ge -6). In exams an outside subtraction shifts the range downward.

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

(\sqrt{x-1}\ge 0), so (2-\sqrt{x-1}\le 2) and it can go down without bound. In exams a negative sign reverses the direction of the range.

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

The denominator (x^2+1) has minimum value (1), so the maximum output is (1) and (0) is not reached. In exams (0) may be a limit, not a value, for reciprocals.

Correct answer: A. ( (3,\frac{7}{2}])

The range of (\frac{2}{x^2+4}) is ( (0,\frac{1}{2}]), so adding (3) gives ( (3,\frac{7}{2}]). In exams apply the vertical shift to the range.

Correct answer: A. ( -3)

The direct answer is option A, -3. To find the minimum of a quadratic, complete the square: x^2-4x+1=(x^2-4x+4)-3=(x-2)^2-3. A square is always at least zero, so (x-2)^2≥0. Its smallest possible value is 0, reached when x=2. Therefore f(x) is at least -3, and the minimum value is -3. Option A is correct. Option B, 1, is the constant term of the original expression, but it is not the minimum. Option C, 3, ignores the negative adjustment produced while completing the square. Option D, 5, is larger than the actual minimum and cannot be obtained as the least value because the square term can be zero. The important idea is that a positive square plus a constant has its minimum when the square becomes zero. Exam cue: rewrite every quadratic in the form (x-a)^2+b; its minimum is b when the coefficient of the square is positive.

Correct answer: B. (8)

The direct answer is B, 8. A square can never be negative, so \\(x-4\\)^2\\ge0\\). Therefore subtracting it from 8 gives \\(8-(x-4)^2\\le8\\). Equality, and hence the maximum, occurs when \\(x-4=0\\), so \\(x=4\\) and \\(f(4)=8\\). Option A, 4, is the x-value where the maximum occurs, not the maximum function value. Option B, 8, is correct because it is the greatest possible output. Option C, 12, is impossible because a nonnegative square cannot make the expression larger than 8. Option D, 0, can occur when \\((x-4)^2=8\\), but it is not the maximum. Memory cue: in \\(a-(x-h)^2\\), the maximum is \\(a\\) at \\(x=h\\).

Correct answer: A. ([0,9])

The interval ([-3,2]) contains (0), so the minimum is (0) and the maximum is (9). In exams check both endpoints and the critical point.

Correct answer: A. ([4,25])

On ([2,5]), (x^2) is increasing, so endpoint values are (4) and (25). In exams include endpoints for a closed interval.

Correct answer: A. ((-3,7])

(2x+1) is an increasing linear function, so the endpoints become (-3) and (7). In exams keep an open endpoint open.

Correct answer: A. ([-10,2])

The function is decreasing, so at (x=0) it gives (2) and at (x=4) it gives (-10). In exams the order may reverse for a decreasing function.

Correct answer: A. ([2,5])

On a negative interval, (|x|) makes values positive. In exams the modulus range of ([-5,-2]) is ([2,5]).

Correct answer: A. ([0,3])

The domain contains (x=1), so the minimum is (0) and the maximum is (3) at an endpoint. In exams always check the zero point of modulus.

Correct answer: A. ([2,4])

(\sqrt{x}) is increasing, so endpoint values are (\sqrt{4}=2) and (\sqrt{16}=4). In exams use monotonicity.

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

On positive (x), (\frac{1}{x}) decreases, so the range is ([\frac{1}{4},1]). In exams reverse the order of endpoint values for a decreasing reciprocal.

Correct answer: A. (\mathbb{R}\setminus{0})

(\frac{1}{x-2}) can never be (0). In exams a horizontal shift keeps (0) excluded from the reciprocal range.

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

(\frac{3}{x+1}) is never (0), so the output (-4) is not obtained. In exams a vertical shift changes the excluded range value.

Correct answer: A. ([0,2])

On the domain, (0\le 4-x^2\le 4), so (0\le f(x)\le 2). In exams the range of a square root is always non-negative.

Correct answer: A. [0, 5]

The governing concept is the range of a square-root function. For f(x) = √(25 − (x − 1)²), the expression under the radical must be non-negative, so (x − 1)² ≤ 25. The quantity (x − 1)² can vary from 0 to 25, making the radicand vary from 25 down to 0. Since a principal square root is always non-negative, f(x) varies from √0 = 0 to √25 = 5, and both endpoints are attained: x = 6 or −4 gives 0, while x = 1 gives 5. Therefore the range is [0, 5]. Option B incorrectly includes negative outputs, option C confuses the input shift with output values, and option D uses the radicand rather than its square root.

Correct answer: A. ([2,\infty))

Inside, (x^2+6x+13=(x+3)^2+4), so the minimum is (4) and the range is ([2,\infty)). In exams complete the square inside first.

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

Since (x^2\ge 0), the output can be (0), but it never reaches (1). In exams (\frac{x^2}{x^2+1}=1-\frac{1}{x^2+1}) is useful.

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

(\frac{x^2+1}{x^2+2}=1-\frac{1}{x^2+2}). The minimum (\frac{1}{2}) is obtained, but (1) is not obtained.

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

The denominator (|x|+2) has minimum value (2) and can grow without bound. So the output is greater than (0) and up to (\frac{1}{2}).

Correct answer: C. (f(x)=\sqrt{x-2})

(\sqrt{x-2}) needs (x-2\ge 0), so all real (x) are not allowed. In exams check the domain restriction in square root options.

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

(\frac{1}{x}) is never (0), so the output cannot be (2). In exams identify the vertical shift of a reciprocal.