Real-Valued Functions, Their Domain and Range

Since (x^2+16\ge16), the square root is at least (4). The inside expression is defined for every real (x).

The governing concept is the domain restriction of a rational function: its denominator must not be zero. Here x^2-4=(x-2)(x+2). Setting the denominator equal to zero gives x=2 or x=-2. These two inputs are excluded because division by zero is undefined. Every other real number makes the denominator non-zero, so the domain is R\{-2,2}, exactly as stated in option A. Option B removes 0 even though the denominator at 0 is -4 and the function is defined there. Option C incorrectly retains only the interval between the zeros, while option D excludes the interval and incorrectly includes the endpoints, where the denominator is zero. Thus A is the only correct statement.

(\sqrt{x-3}=5) gives (x-3=25), so (x=28). In exams check the answer in the original equation after squaring.

Substituting \(f(x)=9\) gives \(|x+6|=9\). A modulus equation has two cases: \(x+6=9\), which gives \(x=3\), and \(x+6=-9\), which gives \(x=-15\). Hence both values are possible. Option D considers only the positive branch and misses the negative one. Exam tip: for \(|A|=k\) with \(k>0\), always solve both \(A=k\) and \(A=-k\).

Substituting \(x=-2\) in the function gives \(f(-2)=(-2)^2+3(-2)-2=4-6-2=-4\). Therefore, option A is correct. Option B may result from overlooking the negative signs in the expression. In exams, use brackets when substituting a negative value.

(f(1)=\frac{1-3}{1+2}=-\frac{2}{3}). In exams handle the numerator and denominator separately with care.

Substitute \(x=-5\): \(f(-5)=\sqrt{(-5)^2-16}=\sqrt{25-16}=\sqrt{9}=3\).

Since (x^2\ge0), (x^2-12\ge-12). In exams remember the minimum value of the square function is (0).

In the given notation, the second set is the codomain (\mathbb{R}), while actual outputs are ([7,\infty)). In exams keep range and codomain separate.

For the given inputs, (f(2)=1), (f(5)=2), (f(10)=3), and (f(17)=4). In exams use only the given inputs in a finite domain.

The expression inside the square root must satisfy (\frac{x+3}{x-1}\ge0) and (x\ne1). A sign table gives ((-\infty,-3]\cup(1,\infty)).

The two square roots give (4\le x\le10) and the denominator gives (x\ne6). In exams take the intersection of all conditions.

The expression inside the square root must satisfy (\frac{x-1}{x-7}\ge0) and (x\ne7). A sign table gives ((-\infty,1]\cup(7,\infty)).

The square root is in the denominator, so ((x+2)(5-x)>0) is required. In exams do not include equality for a denominator radical.

The inside expression is (x^2+4x-12=(x+6)(x-2)) and it must be (\ge0). In exams choose the outer intervals for an upward quadratic.

The square root needs (18-2x-x^2\ge0). This is a downward quadratic, so the closed interval between the roots is the domain.

(x^2-10x+21=(x-5)^2-4), so the minimum is (-4). In exams complete the square to find the range of a quadratic.

Since ((x-5)^2\ge0), (12-(x-5)^2\le12). In exams check the maximum value of a downward-opening parabola.

The direct answer is option A: the range is [2,∞). Start with the expression inside the square root: x²+8x+20. Complete the square: x²+8x+20=(x+4)²+4. A square such as (x+4)² is always at least 0, so the expression inside the root is at least 4. Its smallest value is 4, reached at x=−4. Therefore the smallest function value is √4=2. As x moves far from −4, the square becomes larger without limit, so the function can take every value from 2 upward. Option A is correct. Option B, [20,∞), forgets that the square root changes the minimum 4 into 2. Option C, [0,∞), is too broad because values between 0 and 2 cannot occur. Option D, all real numbers, is wrong because a real square root cannot be negative. Memory cue: complete the square, find the minimum inside the root, then take its square root.

Inside the square root must be non-negative: \(49-(x+3)^2\ge 0\Rightarrow (x+3)^2\le 49\). Then \(49-(x+3)^2\) can take any value from \(0\) up to \(49\). Taking square roots, \(f(x)\) takes any value from \(\sqrt{0}=0\) to \(\sqrt{49}=7\). Therefore the range is \([0,7]\).

The denominator (x^2-6x+13=(x-3)^2+4), whose minimum value is (4). So the output is greater than (0) and up to (\frac{1}{4}).

The direct answer is option A: the range is (−5,−7/2]. Put t=x². Since x²≥0, t≥0, so x²+2≥2. Consequently 0<3/(x²+2)≤3/2. The largest value 3/2 occurs when x=0. The fraction never equals 0, although it approaches 0 as |x| becomes very large. Subtracting 5 shifts every value left by 5, giving −5< f(x)≤−5+3/2=−7/2. Thus the range is (−5,−7/2]. Option A correctly excludes −5 and includes −7/2. Option B wrongly includes −5, which is impossible for a finite x. Option C is the range of the fraction before subtracting 5, not of the complete function. Option D is wrong because the function is always greater than −5 and never reaches values below it. Exam cue: a positive reciprocal has an open lower endpoint when its denominator can grow without limit.

(\frac{x^2+3}{x^2+7}=1-\frac{4}{x^2+7}). At (x=0), (\frac{3}{7}) is obtained and (1) is never obtained.

Rewrite the expression as f(x) = (x²+1+4)/(x²+1) = 1 + 4/(x²+1). For every real x, x² is non-negative, so x²+1 is at least 1 and is always positive. Consequently, 4/(x²+1) is positive and cannot exceed 4. Its largest value, 4, occurs when x=0, giving f(0)=5; hence 5 is included. As |x| increases without bound, 4/(x²+1) approaches 0, so f(x) approaches 1. However, the fraction is never exactly zero for any real finite x, so 1 is not attained. Therefore the range is (1,5], making option A correct. Option B incorrectly includes the unattainable value 1, while C and D do not give the correct lower and upper bounds.

The denominator (|x+4|+3) has minimum value (3) and can grow without bound. Hence the range is ( (0,\frac{1}{3}]).