Real-Valued Functions, Their Domain and Range

In a real valued function, the values are real numbers. The domain (A) can be any suitable set.

The range of a function is the set of output values that the function actually produces when every permitted input from its domain is used. It is therefore a subset of the codomain in the usual definition. For example, if f(x) = x² with domain ℝ and codomain ℝ, the outputs can never be negative, so the range is [0, ∞), even though the codomain is all real numbers. The domain concerns allowed inputs, while the range concerns obtained outputs. Hence option B is correct. Option A describes the domain, option C merely names the codomain rather than defining the range, and option D is not a mathematical definition.

The domain is the set of values that can be used as (x) in the function. While finding it, check denominator and square-root conditions.

The radicand must be non-negative for a real value, so \(x+2\ge 0\). However, the square root is in the denominator and cannot be zero; therefore, \(x+2>0\), or \(x>-2\). Hence the domain is \((-2,\infty)\). Option A is incorrect because it includes \(x=-2\), which makes the denominator zero. Exam tip: when a square root occurs in the denominator, its radicand must be strictly positive.

\(x^2\) is defined and real for every real \(x\). In contrast, \(\sqrt{x}\) needs \(x\geq0\), and \(1/x\) is undefined at \(x=0\). Exam tip: first check the radicand and any denominator when identifying domains.

For a rational function, the denominator must not be zero. Here the denominator is \(x^2+1\). Since \(x^2\geq 0\) for every real \(x\), we have \(x^2+1\geq 1\), so it can never be zero. Therefore, the function is defined for every real number and its domain is \(\mathbb{R}\). The closest distractor is B, but at \(x=0\) the denominator equals \(1\), so zero must not be excluded. Exam tip: To find the domain, reject only those values that make the denominator zero.

Evaluate the function at every element of A: f(0)=1, f(1)=3, f(2)=5, and f(3)=7. Therefore, the range is {1,3,5,7}. Remember that the range contains the output values, whereas the elements of A form the domain.

For inputs −2, −1, 0 and 1, the function values are 5, 2, 1 and 2, respectively. The distinct outputs are therefore {1, 2, 5}, which is the range. Repeated outputs are listed only once.

Substituting \(x=2\), we get \(f(2)=|2-6|=|-4|=4\). Therefore, 4 is correct. The closest distractor is \(-4\), but an absolute value represents a non-negative distance from zero. Exam tip: substitute the given input first, then evaluate the modulus.

For every real \(x\), \(x^2\ge 0\), so \(f(x)=3x^2\ge 0\). Also, when \(x=0\), \(f(0)=0\), so zero is included in the range. Therefore, the range is \([0,\infty)\). Option A is incorrect because it excludes zero, while option C includes negative values that the function cannot produce. Exam tip: For a squared function, first identify its minimum value.

\(x^2+1\) is a polynomial, so it is defined and real for every real \(x\). In contrast, \(\sqrt{x-3}\) needs \(x\ge3\). Exam tip: first check restrictions from roots, denominators, and logarithms.

For every real x, x² ≥ 0, so f(x) = 4x² − 1 ≥ −1. The value −1 is attained at x = 0, and the function grows without bound as |x| increases. Therefore, its range is [−1, ∞).

A real square root is defined only when its radicand is non-negative. Thus, \(3x+12\ge 0\), which gives \(x\ge -4\). Therefore, the domain is \([-4,\infty)\). Option B uses the wrong direction of the inequality, and \(-4\) must be included because the square root is zero there. Exam tip: For a square-root function, set the expression under the root greater than or equal to zero.

The function f(x) = (2x + 1)/3 is a non-constant linear function with non-zero slope 2/3. A non-constant linear function defined for every real x can produce every real output. Algebraically, let y = (2x + 1)/3. Then 3y = 2x + 1, so x = (3y − 1)/2. For every real y, this formula gives a real x, proving that every real number occurs as an output. Thus the range is ℝ, so option B is correct. Options A and D incorrectly impose one-sided bounds, while option C claims the function has only one output, which is impossible for a linear function with non-zero slope.

This is a constant function, so for every real \(x\), the output is always \(-4\). Therefore, its range is the singleton set \(\{-4\}\). Options B and D are intervals, but this function never produces any value other than \(-4\). Exam tip: the range of a constant function is the set containing its single constant value.

For \(x\in[0,9]\), the minimum value of \(\sqrt{x}\) is \(\sqrt{0}=0\), and the maximum value is \(\sqrt{9}=3\). Therefore, the range is \([0,3]\). Option B is the domain of \(x\), not the range of \(f(x)\). Exam tip: For a square-root function on a closed interval, evaluate the function at the interval's endpoints to find the range.

The interval ([-2,3]) includes (0), so the minimum value is (0). The largest square is (3^2=9), so the range is ([0,9]).

The linear function (f(x)=x+5) is defined for every real (x). In exams remember that a linear function usually has domain (\mathbb{R}).

For every real (x), (x^2 \ge 0). In exams remember that the square function has minimum value (0).

For a square root, (x-2 \ge 0), so (x \ge 2). In exams keep the expression under the square root ( \ge 0).

A rational function is defined only when its denominator is not zero, because division by zero has no meaning. Here the denominator is the expression x-3. Therefore, every real number is allowed except the number that makes x-3 equal to zero. The excluded value is 3, so option C follows.

To check it directly, solve x-3=0. Adding 3 to both sides gives x=3. For any other value, the denominator is nonzero and the function has a real value. It is not necessary to exclude 0, 1, or -3, because substituting any of them leaves a nonzero denominator. Thus the domain is all real numbers except 3.

For the square root, (5-x \ge 0), so (x \le 5). In exams handle the inequality sign carefully while solving.

For every real \(x\), \(x^2\geq 0\). Hence \(f(x)=x^2+1\geq 1\). At \(x=0\), \(f(0)=1\), so 1 is the minimum value, and every larger value is possible. Thus, the range is \([1,\infty)\). Option B is wrong because the function can never have the value 0. Exam tip: find the minimum of \(x^2\), then include any constant vertical shift.

A modulus function is defined for every real (x). In exams modulus does not usually create a domain restriction.

A linear function with non-zero slope can produce all real outputs. In exams (ax+b) has range (\mathbb{R}) when (a \ne 0).