Real-Valued Functions, Their Domain and Range

The square root needs (\frac{9-x^2}{x^2-4}\ge 0) and (x\ne\pm2). A sign check gives ([-3,-2)\cup(2,3]).

The square root needs (\frac{x^2-1}{4-x^2}\ge 0) and (x\ne\pm2). A sign check gives ((-2,-1]\cup[1,2)).

The function is (\sqrt{\frac{x+3}{2-x}}), where the fraction takes all values in ([0,\infty)). Hence the square root range is also ([0,\infty)).

The square root needs (1-\frac{1}{x^2}\ge 0) and (x\ne 0). This gives (x^2\ge 1), so (|x|\ge 1).

On the domain, (\frac{1}{x^2}\in(0,1]), so the inside value lies in ([0,1)). After square root, the range remains ([0,1)).

Since \(x^2+1\ge1\), the denominator never becomes zero, so the domain is \(\mathbb R\). Its reciprocal is positive and at most 1, giving range \((0,1]\). Exam tip: check denominator zeros first.

The square root needs (\log_{\frac{1}{4}}(9-x)\ge 0) and (9-x>0). Since the base (\frac{1}{4}<1), (0<9-x\le 1), so the domain is ([8,9)).

Since (\sqrt{x^2+4}\ge 2), the denominator is at least (5), so the maximum value is (\frac{2}{5}). As (|x|\to\infty), the value approaches (0) but never becomes (0).

For a real square root, the expression inside the radical cannot be negative. Hence \(x+5\geq 0\), so \(x\geq -5\). Therefore, the domain is \([-5,\infty)\). Option B excludes \(-5\), but \(f(-5)=\sqrt{0}=0\) is defined. Exam tip: For a square-root function, set the radicand greater than or equal to zero to determine its domain.

\(e^x\) is defined for every real \(x\) and is always positive. As \(x\to-\infty\), \(e^x\to0\), but it never equals 0, so no minimum is attained. In contrast, \(x^2\) and \(|x|\) attain 0. Exam tip: distinguish a limit from an attained value.

\((x+4)^2\ge 0\) for all real \(x\). Since \(y=-(x+4)^2+9\), the term \(-(x+4)^2\le 0\). Therefore \(y\le 9\). The maximum value \(9\) occurs when \((x+4)^2=0\Rightarrow x=-4\). As \((x+4)^2\) grows, \(y\) decreases without bound. Hence the range is \(( -\infty,9 ]\).

For a real square root, the radicand must be non-negative: 4 − x ≥ 0. Solving gives x ≤ 4, so the domain is (−∞, 4].

The denominator of a rational function cannot be zero. Here the denominator is \(x-1\), and solving \(x-1=0\) gives \(x=1\). Therefore, \(x=1\) is excluded from the domain; the other listed values do not make the denominator zero. Exam tip: to find the domain of a rational function, exclude the values that make its denominator zero.

When \(x>2\), both \(x+2\) and \(x-2\) are positive. Hence, \(|x+2|=x+2\) and \(|x-2|=x-2\), so \(y=(x+2)-(x-2)=4\). The option \(2x\) results from mishandling the subtraction of the second modulus term. Exam tip: determine the sign of each expression inside an absolute value before removing the modulus.

The function represents the sum of the distances of \(x\) from 1 and from \(-1\). By the triangle inequality, \(|x-1|+|x+1|\geq |(x-1)-(x+1)|=2\). For \(-1\leq x\leq 1\), this sum is exactly 2, and outside this interval it increases above 2. Hence the range is \([2,\infty)\). Option B incorrectly assumes that the minimum can be 0, although the two points are 2 units apart. Exam tip: For a sum of absolute-value terms, check the interval between their critical points to find the minimum.

For this function, \(x\neq 0\), so \(|x|>0\) and therefore \(y=\frac{1}{|x|}>0\). As \(|x|\) becomes very large, \(y\) approaches zero but never equals it; as \(|x|\) approaches zero, \(y\) can increase without bound. Hence the range is \((0,\infty)\). Option B is incorrect because it includes \(0\). Exam tip: the reciprocal of a positive number is always positive, so negative values cannot belong to the range.

Since \(x^2\ge 0\), we have \(x^2+1>0\) for every real \(x\), so its square root is always real. In option B, the radicand becomes negative when \(|x|>1\). Exam tip: for square-root functions, check that the radicand is always \(\ge0\).

Because |x − 5| ≥ 0, the function is always at least 2. It reaches 2 when x = 5, and |x − 5| can take every non-negative value. Therefore, the range is [2, ∞).

For a real-valued square root, the expression under the radical must be non-negative. Thus, \\(x-6\\ge 0\\), which gives \\(x\\ge 6\\). Therefore, the domain is \\([6,\\infty)\\). Exam tip: For a square-root function, set the radicand \\(\\ge 0\\); 6 is included because \\(\\sqrt{0}\\) is defined.

The radicand must be non-negative: 16 − x² ≥ 0. Hence x² ≤ 16, which gives −4 ≤ x ≤ 4. The endpoints are included because the radicand is zero there, so the domain is [−4, 4].

The square root ensures y ≥ 0. Also, 16 − x² ≤ 16, so y ≤ 4. The value 4 occurs at x = 0, and 0 occurs at x = ±4. Thus the range is [0, 4].

The denominator of a rational function cannot be zero because division by zero is undefined. Setting \(x+2=0\) gives \(x=-2\), so this value is excluded from the domain. For the other given values, the denominator is non-zero, so they belong to the domain. Exam tip: To find excluded values of a rational function, set its denominator equal to zero.

When \(x=-3\), we get \(x+3=0\). The signum function \(\operatorname{sgn}(t)\) has value \(0\) at \(t=0\), so \(f(-3)=0\). Exam tip: \(\operatorname{sgn}(t)\) equals \(-1,0,1\) for \(t<0,t=0,t>0\), respectively.

Since |x + 5| ≥ 0, y = 2 − |x + 5| ≤ 2. At x = −5, the function reaches 2. As |x + 5| grows, y decreases without bound, so the range is (−∞, 2].

For \\(\sqrt{1-x^2}\\), the radicand must satisfy \\(1-x^2\ge0\\), giving \\(-1\le x\le1\\). A square root is non-negative, and its maximum is 1 at \\(x=0\\). Thus the range is \\([0,1]\\). Exam tip: first impose the radicand condition.