Real-Valued Functions, Their Domain and Range

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).

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

(\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.

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.

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.

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.

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\\).

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.

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

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

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.

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

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.

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

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.

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

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

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.

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.

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.

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.

(\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.

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}).

(\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.

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