Real-Valued Functions, Their Domain and Range

Since (x^2+9\ge 9), the square root is at least (3) and the domain is all real numbers. In exams check the minimum value inside.

The denominator (x^2-1) becomes (0) when (x=\pm 1). In exams remove denominator zero values in rational functions.

(\sqrt{x+2}=3) gives (x+2=9) and (x=7). In exams keep the domain condition valid after squaring.

Substituting \(f(x)=6\) gives \(|x-4|=6\). A modulus equation has two cases: \(x-4=6\), giving \(x=10\), and \(x-4=-6\), giving \(x=-2\). Hence the correct set is \(\{-2,10\}\). In option B, \(x=2\) gives \(|2-4|=2\), not 6. Exam tip: for \(|A|=k\) with \(k>0\), always solve both \(A=k\) and \(A=-k\).

For the square root, \(3-x\geq0\), so \(x\leq3\). Also, the denominator cannot be zero, so \(x=1\) is excluded. In exams, check radical and denominator restrictions separately.

Substituting \(x=4\) in the given function gives \(f(4)=\frac{4+2}{4-1}=\frac{6}{3}=2\). Therefore, option A is correct. The denominator is \(x-1\), so the domain excludes \(x=1\); \(x=4\) is valid. Exam tip: in a rational function, first check that the denominator is non-zero, then substitute the given value.

The radicand \(x-2\) must be non-negative, but the denominator \(\sqrt{x-2}\) cannot be zero. Hence \(x-2>0\), so \(x>2\). Option B wrongly includes \(x=2\), where the denominator is zero. Exam tip: a square root in a denominator requires a strict inequality.

Direct answer: A, [-9,∞). For every real x, x^2 is at least 0. Thus x^2-9 is at least -9. The value -9 is attained at x=0, and every value greater than -9 is possible: if y≥-9, choose x=√(y+9) or its negative. Option A is therefore the actual range. Option B, all real numbers, is the codomain but not the actual output set; negative values below -9 cannot occur. Option C [0,∞) describes x^2 before subtracting 9. Option D (-∞,-9] incorrectly reverses the inequality and gives only values at or below -9. Memory cue: complete the square or use x^2≥0; distinguish codomain from range.

The second set in the function notation is the codomain, while actual outputs are ([4,\infty)). In exams identify codomain from arrow notation.

The values are (f(-2)=2), (f(0)=0), and (f(3)=12). In exams find the image of every input in a finite domain.

For (x=1,2,4,5), the outputs are (0,1,\sqrt{3},2). In exams use only the given inputs in a set domain.

The expression inside the square root must satisfy (\frac{x-2}{x+1}\ge 0) and (x\ne -1). A sign table gives ((-\infty,-1)\cup[2,\infty)).

For a real-valued function, every square-root radicand must be nonnegative, and every denominator must be nonzero. The radical condition is x + 1 ≥ 0, which gives x ≥ −1. The rational term requires x − 2 ≠ 0, so x ≠ 2. Combining these restrictions gives the domain [−1, ∞) excluding 2, written [−1, ∞) \ {2}. The endpoint −1 is included because √0 is defined, whereas 2 is removed because division by zero is undefined. Option A correctly includes −1 and excludes 2. Option B wrongly excludes −1; option C excludes many valid values; option D includes values less than −1, where the square root is not real.

The square root needs (5-x\ge 0) and the denominator needs (x+3\ne 0). In exams take the intersection of both conditions.

For \(f(x)=\sqrt{x^2-9}\) to be real, we need \(x^2-9\ge 0\). That means \(x^2\ge 9\), so \(x\le -3\) or \(x\ge 3\). Therefore the domain is \(( -\infty,-3]\cup[3,\infty)\).

The direct answer is A, \\((-4,4)\\). A real square root requires its radicand to be nonnegative, but here the square root is in the denominator, so the radicand must actually be positive: \\(16-x^2>0\\). This gives \\(x^2<16\\), hence \\(-4<x<4\\). The endpoints cannot be included because at x=4 or x=-4 the denominator becomes zero, and division by zero is undefined. Option A is correct. Option B includes the endpoints, so it incorrectly allows denominator zero. Option C describes values outside the interval, where \\(16-x^2<0\\) and the square root is not real. Option D excludes only the endpoints but includes outside values, which also fail the real-square-root condition. Exam cue: a square root in a denominator needs a strictly positive radicand.

The denominator is (x^2-16=(x-4)(x+4)), so (x=\pm4) are excluded. In exams apply restrictions from the original denominator before cancellation.

The denominator (x^2-x-20=(x-5)(x+4)), so (x\ne5) and (x\ne-4). In exams factorize the denominator.

The two conditions (x-2\ge0) and (9-x\ge0) together give (2\le x\le9). In exams take the common interval for two square roots.

The denominator (x^2-4x+5=(x-2)^2+1) is never (0). In exams complete the square to check the minimum denominator value.

The square root needs (15-5x\ge0), giving (x\le3). In exams the inequality reverses when dividing by a negative coefficient.

The square root is real only when its radicand is greater than or equal to zero. Thus we solve (x + 1)(x − 5) ≥ 0. The critical values are −1 and 5. On a sign chart, both factors are negative when x < −1, making their product positive; between −1 and 5, one factor is negative and the other positive, making the product negative; and for x > 5, both are positive, making the product positive. The endpoints are included because the radicand is zero there. Hence the domain is (−∞, −1] ∪ [5, ∞), option A. Option B incorrectly selects the interval where the product is negative.

(x^2+8x+18=(x+4)^2+2), so the minimum is (2). In exams completing the square is useful for the range of a quadratic.

Since ((x-3)^2\ge0), (7-(x-3)^2\le7). In exams identify the maximum value of a downward-opening parabola.

The key property of an absolute value is |x + 4| ≥ 0 for every real x. Therefore f(x) = |x + 4| − 5 is always at least −5. Equality occurs when x + 4 = 0, or x = −4, so the minimum value −5 is actually attained. As x moves away from −4, |x + 4| can take every nonnegative value, and consequently f(x) can take every value from −5 upward. Thus the range is [−5, ∞), making option A correct. Option B incorrectly ignores the downward shift by 5; option C reverses the direction, and option D includes values less than −5 that the absolute-value expression can never produce.