Real-Valued Functions, Their Domain and Range

The radicand must be non-negative: 2 − |x − 1| ≥ 0. Thus |x − 1| ≤ 2, which gives −1 ≤ x ≤ 3. Both endpoints are included because the radicand is zero there.

First impose the real-valued condition for the square root: 2 − |x − 1| ≥ 0, or |x − 1| ≤ 2. The expression inside the radical therefore varies from 0 to 2. Its maximum 2 occurs at x = 1, while the minimum 0 occurs at x = −1 and x = 3. Because the square-root function is nonnegative and increasing on nonnegative inputs, taking the square root changes the interval [0, 2] into [0, √2]. Both endpoints are actually attained, so they must be included. Hence option A is correct. Option B confuses the radicand with the function value, option C incorrectly permits negative outputs, and option D excludes attainable endpoint values.

Because the square root is in the denominator, its radicand must be strictly positive: |x + 2| − 5 > 0. Hence |x + 2| > 5. Splitting the absolute-value inequality gives x + 2 > 5 or x + 2 < −5, so x > 3 or x < −7. The boundary values x = 3 and x = −7 make the radicand zero, producing a zero denominator, and are therefore excluded. All values outside these two boundaries make the radicand positive. Thus the domain is (−∞, −7) ∪ (3, ∞), which is option A. Option B incorrectly includes the zero-denominator boundaries, C gives the region where the radicand is not positive, and D fails to exclude the whole interval between the boundaries.

From (y=\frac{4x-1}{2x+5}), we get (x=\frac{-1-5y}{2y-4}), so (y\ne2). For a linear fractional function, the missing value is often the ratio of leading coefficients.

For the square root, ((x-2)(7-x)\ge0) gives (x\in[2,7]), and the denominator removes (x=3). Taking the intersection of all restrictions is essential.

The denominator is (x^2-4x+8=(x-2)^2+4), so the fraction has maximum (\frac{1}{2}) and never attains (0). Hence the total range is ((1,\frac{3}{2}]).

The expression inside the square root must satisfy (3x-12\ge 0), so (x\ge 4). In exams, always keep the radicand non-negative.

Here (25-x^2\ge 0), so (x^2\le 25) and (x\in[-5,5]). In such questions, solve the inequality carefully.

For real-valued outputs, \(x-2\ge 0\) is required because it is under a square root. Since the square root is in the denominator, it cannot be zero; therefore \(x-2>0\), giving \(x>2\). Hence the domain is \((2,\infty)\). Option B is incorrect because at \(x=2\), the denominator becomes zero. Exam tip: when a square root occurs in the denominator, its radicand must be strictly positive.

The direct answer is option A: the domain is all real numbers except −3 and 3, written ℝ\{−3,3}. A rational function is defined wherever its denominator is not zero. Here the denominator is x²−9. Set it equal to zero: x²−9=0, so x²=9 and x=3 or x=−3. These two values must be excluded. The numerator x+1 causes no restriction because it is allowed to be zero; a zero numerator simply makes the function value zero, provided the denominator is nonzero. Option A correctly removes both forbidden values. Option B removes only 3 and wrongly allows −3, where the denominator is zero. Option C removes only −3 and wrongly allows 3. Option D lists only the excluded values instead of all permitted real values. Every other real number is allowed, including x=−1, where the numerator is zero and the denominator is −8. Memory cue: for a fraction, solve denominator=0 and delete those values only; never delete values merely because the numerator is zero.

For the square root (x+1\ge 0) and for the denominator (x\ne 4) are required. Thus the intersection is ([-1,\infty)-{4}).

For every real \(x\), \((x-2)^2\ge 0\), so \(f(x)\ge 5\). Equality holds at \(x=2\), where \(f(2)=5\); therefore, 5 is included and the range is \([5,\infty)\).

For every real \(x\), \((x+3)^2\geq 0\). Hence \(f(x)=7-(x+3)^2\leq 7\). At \(x=-3\), the squared term is zero, so \(f(-3)=7\); thus 7 is the maximum value and is included in the range. As \(|x+3|\) increases, \(f(x)\) decreases without bound. Therefore, the range is \(( -\infty, 7]\). The interval \([7,\infty)\) is incorrect because the function cannot take a value greater than 7. Exam tip: a parabola of the form \(a-(x-h)^2\) has maximum value \(a\).

For a rational function, the denominator cannot be zero. Here, \(x-3\ne 0\), so \(x\ne 3\). Therefore, every real number except 3 belongs to the domain, giving \(\mathbb{R}\setminus\{3\}\). Option B incorrectly excludes \(-3\), and option D includes the undefined value 3. Exam tip: To find the domain of a rational function, first set its denominator equal to zero and exclude the resulting values.

If (y=\frac{2x+1}{x-3}), then (x=\frac{3y+1}{y-2}), so (y\ne 2). Hence the range is (\mathbb{R}-{2}).

For every real x, \(x^2\ge 0\), so \(x^2+4\ge 4\). Hence \(0<\frac{1}{x^2+4}\le\frac14\). The maximum value \(\frac14\) occurs at \(x=0\), while 0 is never attained; therefore, the range is \((0,\frac14]\). Exam tip: for a positive denominator, the reciprocal is largest when the denominator is smallest.

For a real-valued square root, the expression inside the radical must be non-negative. Hence \(9-(x-1)^2\ge 0\), so \((x-1)^2\le 9\). This gives \(-3\le x-1\le 3\), or \(-2\le x\le 4\). Therefore, the domain is \([-2,4]\). Option C contains values for which the expression inside the square root is negative. Exam tip: To find the domain of a square-root function, set its radicand greater than or equal to zero.

Since \(|x-4|\ge 0\), we have \(f(x)=|x-4|+2\ge 2\). At \(x=4\), \(|x-4|=0\), so \(f(4)=2\); therefore, the value 2 is attained. As \(|x-4|\) increases, all values greater than 2 are also obtained, giving the range \([2,\infty)\). Option B is incorrect because it excludes the attained minimum value 2. Exam tip: For a function of the form \(|x-a|+b\), the range is \([b,\infty)\).

For every real \(x\), \(|x|\ge 0\), so \(|x|+1\ge 1\), which gives \(0<f(x)\le 1\). At \(x=0\), the function equals 1. As \(|x|\) increases, the function approaches 0 but never becomes 0. Hence, the range is \((0,1]\). Option B is incorrect because 0 is not attained. Exam tip: For a reciprocal function, identify the minimum and possible limiting values of the denominator to determine the range.

To determine the range, set t = x², so t ≥ 0. Then f = (t − 1)/(t + 1). Since t + 1 is positive, f + 1 = 2t/(t + 1) ≥ 0, which gives f ≥ −1. The value −1 is attained when t = 0, that is, x = 0. Also, f = 1 would require t − 1 = t + 1, an impossibility because it would give −1 = 1. More directly, 1 − f = 2/(t + 1) > 0, so f < 1 for every real x. As t increases from 0 without bound, f increases from −1 toward 1, reaching every intermediate value but never reaching 1. Therefore the range is [−1, 1), so option A is correct. The other choices either omit −1 or include the unattainable endpoint 1.

For a real square root, the expression inside the radical must be non-negative. Hence, \(x^2-16\geq 0\), or \((x-4)(x+4)\geq 0\). This inequality holds when \(x\leq -4\) or \(x\geq 4\). Therefore, the domain is \(( -\infty,-4]\cup[4,\infty)\). Options B and C contain values between the roots, for which \(x^2-16<0\). Option D is incorrect because at \(x=-4\) and \(x=4\), the square root is 0. Exam tip: For a real square-root function, set the radicand greater than or equal to zero.

For \(f(x)=\sqrt{x^2-16}\) to be defined, we need \(x^2-16\ge 0\Rightarrow x^2\ge 16\Rightarrow |x|\ge 4\). The square root is always nonnegative, so \(f(x)\ge 0\). As \(|x|\) increases, \(x^2-16\) increases without bound, so \(\sqrt{x^2-16}\) also has no upper limit. Therefore the range is \([0,\infty)\).

The argument of a logarithm must be strictly positive, so \(x^2-5x+6>0\). Factoring gives \((x-2)(x-3)>0\), which holds when \(x<2\) or \(x>3\). Therefore, the domain is \((-\infty,2)\cup(3,\infty)\). Option B is incorrect because the logarithm's argument is negative between 2 and 3; the endpoints are also excluded because the argument becomes zero there. Exam tip: For \(\log_a(g(x))\), always impose \(g(x)>0\), not merely \(g(x)\ne0\).

For a logarithmic function, its argument must be positive. Thus, \(4-x>0\), which gives \(x<4\). Therefore, the domain is \(( -\infty,4)\). Option C wrongly includes \(4\); at \(x=4\), the argument is \(0\), and \(\log 0\) is undefined. Exam tip: In logarithmic domain questions, first set the expression inside the log greater than zero.

The condition is (\frac{x-1}{x+2}\ge 0) and (x\ne -2). A sign chart gives ((-\infty,-2)\cup[1,\infty)).