Real-Valued Functions, Their Domain and Range

For (\sqrt{x}), (x\ge 0), and for the denominator square root, (4-x>0) is needed. Together, (0\le x<4).

The direct answer is option A, \(\mathbb R\setminus\{2\}\). The domain contains all real inputs for which the formula is defined. A fraction is undefined when its denominator is zero. Here the original denominator is \(x-2\), so we require \(x-2\ne0\), which means \(x\ne2\). Thus every real number except 2 belongs to the domain. Although \(x^2-4=(x-2)(x+2)\), cancelling gives \(f(x)=x+2\) only for \(x\ne2\). At \(x=2\), the original expression is \(0/0\), undefined; cancellation cannot restore that missing input. Option A is correct. Option B incorrectly includes 2. Option C incorrectly removes -2 as well, even though the original denominator at -2 is -4 and is non-zero. Option D includes only numbers less than 2 and wrongly omits all numbers greater than 2. Memory cue: for a rational function, inspect the original denominator before cancelling factors.

For a real square root, the expression under the radical must be non-negative. Hence \(|x|-3\ge 0\), so \(|x|\ge 3\). Therefore, \(x\le -3\) or \(x\ge 3\), giving the domain \(( -\infty,-3]\cup[3,\infty)\). Option B wrongly includes values between \(-3\) and \(3\), where \(|x|-3<0\). Exam tip: For real functions involving a square root, first set the radicand greater than or equal to zero.

The denominator must not be zero. Solving \(|x|-2=0\) gives \(|x|=2\), so \(x=-2\) or \(x=2\) makes the function undefined. All other real numbers are allowed; hence the domain is \(\mathbb{R}\setminus\{-2,2\}\).

Let \(t=|x|\). Then \(t\ge 0\) and \(f(x)=\frac{t}{1+t}\). At \(x=0\), we get \(t=0\) and hence \(f(x)=0\). For every finite \(t\), \(\frac{t}{1+t}<1\), so the value 1 is never attained, although the function approaches 1 as \(|x|\) increases. Therefore, the range is \([0,1)\). Option B is incorrect because 0 is included, while option C is incorrect because 1 is excluded. Exam tip: for rational functions, check both limiting values and whether each endpoint is actually attained.

For positive (x), the value is at least (2), and for negative (x), it is at most (-2). Check both cases separately.

The direct answer is A: [0,∞). First simplify the expression: x^2-4x+4=(x-2)^2, so f(x)=√((x-2)^2)=|x-2|. An absolute value can never be negative. Its smallest value is 0, reached when x=2. As x moves away from 2, |x-2| can become any positive number; for example, choosing x=2+t gives f(x)=t for every t≥0. Therefore the range is [0,∞). Option A is correct because it includes zero and every non-negative real value. Option B, (0,∞), wrongly excludes zero, even though f(2)=0. Option C, [2,∞), wrongly says the minimum is 2; the actual minimum is 0. Option D, all real numbers, is wrong because the square-root output is never negative. Remember that √(u^2)=|u|, not simply u, and an absolute value has a V-shaped graph with lowest point on the x-axis. Exam cue: whenever a square root contains a perfect square, simplify it using an absolute value and then identify the least possible output.

The maximum of (\sqrt{16-x^2}) is (4), so the minimum fraction is (\frac{1}{4}). Near the endpoints, the denominator tends to (0).

The greatest integer function gives the largest integer not greater than (x). Since (4) is not included, the value (4) will not occur.

The fractional part is always at least (0) and less than (1). At integer (x), the value is (0).

The expression inside the square root must satisfy (\frac{x-1}{x+2}\ge 0) and (x\ne -2). A sign chart gives (x<-2) or (x\ge 1).

For a real square root, the radicand must be non-negative, and its denominator must be nonzero. The critical points are \(-3\), where the numerator is zero, and \(5\), where the denominator is zero. Sign analysis gives \(\frac{x+3}{5-x}\ge 0\) for \(-3\le x<5\). Thus \(-3\) is included, but \(5\) is excluded, so the domain is \([-3,5)\).

The expression under a square root must be non-negative, so \(x-a\ge 0\), which gives \(x\ge a\). Therefore, the domain of \(f(x)=\sqrt{x-a}\) is \([a,\infty)\). Comparing it with the given domain \([7,\infty)\), we get \(a=7\). Choosing \(-7\) would give the domain \([-7,\infty)\), so it is incorrect. Exam tip: The domain of \(\sqrt{x-c}\) is directly \([c,\infty)\).

The denominator \(x-a\) must not be zero. Thus, the only excluded value from the domain is obtained from \(x-a=0\), which gives \(x=a\). Since the excluded value is \(5\), we have \(a=5\). Choosing \(-5\) would exclude \(-5\), not \(5\). Exam tip: For a reciprocal expression, set its denominator equal to zero to identify the value excluded from the domain.

Put (t=(x+1)^2\ge 0), then (f=\frac{t+4}{t+1}=1+\frac{3}{t+1}). Hence the values are greater than (1) and up to (4).

The direct answer is A: [3,4]. For a square root to be real, its inside must be non-negative: 2x−x²≥0. Rewrite it as 1−(x−1)². Since (x−1)² is at least 0, the expression 1−(x−1)² has maximum 1. On its allowed domain, it has minimum 0, so 0≤2x−x²≤1. Taking square roots gives 0≤√(2x−x²)≤1. Adding 3 throughout gives 3≤f(x)≤4. Thus the range is [3,4]. Option A is correct. Option B [0,1] is the range of the square-root part before adding 3. Option C [2,4] incorrectly includes values below 3. Option D [3,∞) incorrectly allows values larger than 4, although the square-root part never exceeds 1. Memory cue: find the range inside the root first, then apply the outside operations.

Complete the square in the denominator: \(x^2-6x+10=(x-3)^2+1\). Since \((x-3)^2\ge 0\), the denominator has minimum value \(1\), attained at \(x=3\). Therefore, the maximum value of \(f(x)\) is \(\frac{1}{1}=1\). For instance, \(\frac{1}{10}\) is not the maximum; it occurs only for particular values of \(x\). Exam tip: when the denominator is positive, the reciprocal is maximized when the denominator is minimized.

For both square roots to be real, each radicand must be non-negative. From \(x+4\ge 0\), we get \(x\ge -4\), and from \(4-x\ge 0\), we get \(x\le 4\). Their intersection is \([-4,4]\), so option A is correct. Option B wrongly excludes \(-4\) and \(4\); at each endpoint, one square root is zero and the function is still defined. Exam tip: For several square roots, write a non-negative condition for every radicand and intersect the resulting sets.

At the endpoints the value is (2\sqrt{2}), and at (x=0) the maximum is (4). For symmetric radical functions, also check the midpoint.

The argument of a real logarithm must be positive, so 4 − x² > 0. This gives x² < 4, or −2 < x < 2. The endpoints are excluded because the logarithm's argument would be zero.

Each logarithm's argument must be positive. The conditions are x − 1 > 0, so x > 1, and 5 − x > 0, so x < 5. Their intersection is (1, 5); the endpoints are excluded because an argument would be zero.

For every real \(x\), \(|2x-1|\ge 0\), so \(f(x)\ge 3\). At \(x=\frac{1}{2}\), the absolute-value term is zero and \(f(x)=3\), so the minimum is attained. As \(|2x-1|\) can grow without bound, the range is \([3,\infty)\).

The radicand must be non-negative: x² − 1 ≥ 0. Hence x² ≥ 1, so x ≤ −1 or x ≥ 1. Therefore, the domain is (−∞, −1] ∪ [1, ∞), including the endpoints where the radicand is zero.

The direct answer is A: [0,∞). A square root can never be negative. First require x²−1≥0, so x≤−1 or x≥1; this is the domain. At x=1 or x=−1, f(x)=√0=0, so 0 is included. For any y≥0, choose x=√(y²+1), which belongs to the domain; then √(x²−1)=√y²=y. Therefore every non-negative real number occurs, and no negative number can occur. Option A is correct. Option B [1,∞) wrongly excludes the value 0 and all values between 0 and 1. Option C includes negative outputs, impossible for a square root. Option D contains only non-positive values and misses all positive outputs. Memory cue: √ means the principal, non-negative root; check both domain and range.

(|4x|\le x^2+4) because ((x-2)^2\ge 0) and ((x+2)^2\ge 0). Hence values lie from (-1) to (1), and both endpoints occur.