Real-Valued Functions, Their Domain and Range

The minimum value of (|x-1|) is (0), so (f(x)\ge2). In exams the outside added (2) shifts the minimum upward.

(\sqrt{x+9}\ge0), so (\sqrt{x+9}-4\ge-4). In exams apply the vertical shift to the basic square-root range.

(\sqrt{x+2}\ge0), so (6-\sqrt{x+2}\le6) and can decrease without bound. In exams a negative sign changes the direction of the range.

The denominator (x^2+9) has minimum value (9), so the maximum output is (\frac{1}{9}). (0) is only a limiting value, not an attained value.

Direct answer: The range is \((-1,-\frac{1}{5}]\), so option A is correct. For every real x, x^2≥0, hence x^2+5≥5. Therefore 0<4/(x^2+5)≤4/5. The value 4/5 occurs at x=0. The fraction can approach 0 when |x| becomes very large, but it never equals 0, so its range is (0,4/5]. Subtracting 1 from every value shifts the interval left by 1, giving (-1, -1+4/5] = (-1,-1/5]. Option A has the correct open lower end and closed upper end. Option B wrongly includes −1, which is impossible. Option C is the range before the subtraction and misses the shift. Option D includes values below −1 that cannot occur. Memory cue: denominator is smallest at x=0, giving the maximum; infinity is approached but never reached.

The governing concept is the minimum value of an upward-opening quadratic function. Complete the square: f(x) = x² + 2x − 8 = (x + 1)² − 9. For every real value of x, the square (x + 1)² is at least zero, and it reaches zero when x = −1. Therefore the least possible value of f(x) is 0 − 9 = −9. The parabola opens upward because the coefficient of x² is positive, so this vertex value is the global minimum. Option A is correct. Option B misses the constant adjustment, option C does not follow from the completed-square form, and option D incorrectly assumes that the minimum must be zero.

\((x+2)^2\ge 0\) for all real \(x\). Since \(f(x)=10-(x+2)^2\), subtracting a nonnegative quantity from \(10\) makes \(f(x)\le 10\). The maximum happens when \((x+2)^2=0\Rightarrow x=-2\). Then \(f(-2)=10-0=10\).

The interval ([-4,1]) contains (0), so the minimum is (0) and the maximum is (16). In exams check the vertex along with endpoints.

Every value in the domain \([-6,-2]\) is negative. For \(f(x)=x^2\), the endpoint values are \(f(-6)=36\) and \(f(-2)=4\). Since squaring takes all intermediate values on this interval, the range is \([4,36]\). Option B is incorrect because obtaining \(0\) would require \(x=0\), which is not in the domain. Exam tip: For a continuous function on a closed interval, check the endpoint values and any relevant extreme values to determine the range.

The governing concept is mapping a domain interval through a real-valued function. Since f(x)=3x−2 has positive slope 3, it is strictly increasing, so the endpoint outputs determine the range. The lower domain endpoint −1 is excluded; its output is f(−1)=3(−1)−2=−5, which must also be excluded. The upper endpoint 5 is included; f(5)=15−2=13, so 13 is included. Hence the range is (−5,13], making option A correct. Option B incorrectly includes −5, option C merely repeats the domain, and option D uses an incorrect lower output. The open and closed brackets are essential because they reflect whether the corresponding domain endpoints are excluded or included.

The function is decreasing, so at (x=1) it gives (2) and as (x\to6) it approaches (-8), not included. In exams reverse endpoint order for a decreasing function.

The domain contains (0), so the minimum is (0) and the largest modulus is (7). In exams check whether the interval contains (0).

For every \(x\in[1,5]\), \(x+2\) is positive, so \(f(x)=x+2\). This function increases on the domain: its minimum is \(f(1)=3\), and its maximum is \(f(5)=7\). Therefore, the range is \([3,7]\).

(\sqrt{x-1}) is increasing, so endpoint values are (0) and (3). In exams include endpoints of a closed interval.

On the positive interval, (\frac{1}{x+1}) decreases. At (x=1) and (x=3), values are (\frac{1}{2}) and (\frac{1}{4}).

(\frac{1}{x+6}) can never be (0). In exams a horizontal shift does not change the excluded range value of a reciprocal.

(\frac{5}{x-2}) is never (0), so output (1) is not obtained. In exams a vertical shift changes the excluded value.

The governing concept is the range of a square-root function. A square root is always non-negative, so f(x) cannot be less than 0. For the expression to be defined, 9-(x+1)^2 must lie between 0 and 9. Its maximum value is 9, attained when x=-1, giving f(-1)=√9=3. Its minimum allowed value is 0, attained when x=-4 or x=2, giving f(x)=0. Every intermediate value from 0 through 3 occurs, so the range is [0,3]. Thus option A is correct. Option B incorrectly permits negative outputs, option C confuses the radicand with its square root, and option D ignores the real-domain restriction.

The governing idea is that a principal square root produces only non-negative outputs. The radicand 36-(x-2)^2 is at most 36 because (x-2)^2 is at least 0. Its maximum 36 occurs at x=2, giving f(2)=√36=6. The radicand becomes 0 at x=-4 and x=8, so the minimum output 0 is also attained. As x varies over the real domain [-4,8], the radicand takes every value from 0 to 36, and the square root consequently takes every value from 0 to 6. Therefore option A, [0,6], is correct. Option B wrongly includes negative outputs, option C confuses the horizontal shift with output values, and option D gives the radicand interval rather than its square-root interval.

Direct answer: A, [3,∞). Rewrite the expression inside the square root: x^2-2x+10=(x-1)^2+9. Since a square is at least 0, the inside is at least 9. The square root is therefore at least √9=3. Equality occurs at x=1, so 3 is included; as |x-1| grows, the expression grows without bound, so every value above 3 is possible. Option A is correct. Option B [10,∞) forgets to take the square root of the minimum 9. Option C [0,∞) is too broad and ignores the positive lower bound. Option D all real numbers is impossible because a principal square root is never negative. Memory cue: complete the square first, then apply the square-root restriction.

(f(x)=1-\frac{4}{x^2+4}), so (0) is attained but (1) is not. In exams writing such rational expressions in (1-) form is useful.

(\frac{x^2+2}{x^2+5}=1-\frac{3}{x^2+5}). At (x=0), the minimum (\frac{2}{5}) is attained and (1) is not attained.

The denominator (|x-3|+1) has minimum value (1) and can grow without bound. Hence the output is greater than (0) and up to (1).

The polynomial (x^2+8) is defined for every real (x). In exams polynomial functions usually have domain (\mathbb{R}).

(\frac{2}{x}) is never (0), so output (-3) cannot occur. In exams identify the vertical shift of a reciprocal function.