Real-Valued Functions, Their Domain and Range

(\sqrt{x^2}) is always non-negative, so it equals (|x|). A common mistake is writing it directly as (x).

A real valued function has output values in the real numbers. The key focus is on the codomain or values.

In \(f:A\to\mathbb{R}\), the set \(A\) contains the input values of the function, so it is called the domain. The set \(\mathbb{R}\) is the codomain, while the range consists only of the values actually produced by the function and is a subset of \(\mathbb{R}\). Exam tip: in function notation, the set before the arrow is the domain and the set after the arrow is the codomain.

The direct answer is option A: codomain. In notation f:A→ℝ, the arrow describes where inputs come from and where outputs are allowed to go. A is the domain, the set of permitted inputs. ℝ is the codomain, the stated target set for outputs; it means the set of real numbers. The actual values that the function really produces are called the range, and the range is always a subset of the codomain, not necessarily the whole codomain. Therefore option A is correct. Option B, domain, is wrong because the domain appears before the arrow, namely A. Option C, only range, is not correct because the symbol ℝ after the arrow specifies the codomain; the range may be smaller depending on the formula. For example, a function may map into ℝ but produce only nonnegative values. Option D, empty set, is wrong because ℝ contains all real numbers and is not empty. Remember the reading cue: in f:A→B, read “from A to B”; A is domain and B is codomain.

((-2)^2=4) and (2^2=4), so only one distinct value is obtained. Repeated values are written once in the range.

The denominator must not be zero. Since x − 2 = 0 at x = 2, exclude 2 from the real numbers. Thus the domain is ℝ ∖ {2}.

For a real square root, the expression inside the radical must be non-negative. Thus, \(2x-6\geq 0\), which gives \(x\geq 3\). Therefore, the domain is \([3,\infty)\). Option B excludes \(3\), but \(f(3)=0\) is defined. Exam tip: For a real square-root function, always set the radicand to be \(\geq 0\).

For a real square root, the radicand must be non-negative: 4 − 2x ≥ 0. Solving gives x ≤ 2, so the domain is (−∞, 2]. The endpoint 2 is included because √0 is defined.

For every real \(x\), \(|x|\geq 0\), so \(f(x)=|x|+2\geq 2\). The minimum value of \(|x|\) is \(0\), attained at \(x=0\); hence the minimum value of the function is \(2\). Since \(|x|\) can take every non-negative value, the range is \([2,\infty)\). Option B omits the added 2. Exam tip: For a modulus function, first identify the minimum value of the modulus expression.

For every real \(x\), \(|x|\geq 0\). Hence \(f(x)=3-|x|\leq 3\), and the maximum value 3 is attained when \(|x|=0\), that is, at \(x=0\). The value 0 occurs when \(|x|=3\), but it is not the maximum. Exam tip: For a function of the form \(a-|x|\), the maximum value is \(a\), attained at \(x=0\).

Substituting \(x=5\) into the function gives \(f(5)=\frac{5+1}{2}=\frac{6}{2}=3\). Therefore, the correct answer is 3. The option 2 results from failing to divide the numerator value 6 by 2. Exam tip: To find a function value, substitute the given input for every occurrence of \(x\) and simplify the resulting expression.

Substitute x = 1 into each term: f(1) = 1² + 2(1) + 1 = 1 + 2 + 1 = 4. Therefore, option A is correct.

For every real \(x\), \((x-1)^2\ge 0\). Its minimum value is \(0\) when \(x=1\), and the function can increase without bound as \(x\) moves away from 1. Therefore, the range is \([0,\infty)\). Option B is incorrect because the minimum value is \(0\), not \(1\). Exam tip: For a squared function, find its minimum or maximum value first to determine the range.

\(f(x)=x^3\) can produce every real value: for any \(y\in\mathbb{R}\), choose \(x=\sqrt[3]{y}\), giving \(f(x)=y\). In contrast, \(x^2\) and \(|x|\) never give negative outputs. Exam tip: an odd-power cubic with real domain has range \(\mathbb{R}\).

For every real \(x\), \(x^2\ge 0\), so \(x^2+4\ge 4>0\). Thus the denominator is never zero, and no real value of \(x\) must be excluded; hence the domain is \(\mathbb{R}\). Options B and C are incorrect because at \(x=2\) and \(x=-2\), the denominator equals 8, not 0. Exam tip: for a fractional function, exclude only those input values that make the denominator zero.

The governing concept is the real-domain restriction for an even square root. For √x to be a real number, its radicand must satisfy x ≥ 0. Multiplying the square root by 2 changes output values but does not introduce any additional restriction on x. Therefore every x in [0, ∞) is allowed, including x = 0, where f(0) = 0. Negative inputs are excluded because the square root of a negative real number is not real. Option B is too narrow because it omits the valid endpoint 0. Option C incorrectly permits negative inputs, while option D contains non-positive values even though only 0 from that set is allowed. Thus the real domain is [0, ∞).

For real x, the function is defined when x ≥ 0, and 2√x is always non-negative. It takes the value 0 at x = 0; moreover, for any y ≥ 0, choosing x = (y/2)² gives f(x) = y. Thus every non-negative value occurs, so the range is [0, ∞).

For \(\frac{1}{\sqrt{x-1}}\), the expression \(\sqrt{x-1}\) must be defined and also cannot be \(0\) because it is in the denominator.

So we need \(x-1>0\Rightarrow x>1\). Therefore the domain is \((1,\infty)\).

For every real \(x\), \(|x+1|\geq 0\). Its minimum value is \(0\), attained when \(x=-1\). Hence, the minimum value of \(f(x)=|x+1|-3\) is \(0-3=-3\). Exam tip: An absolute-value expression has minimum value \(0\) before any constant is added or subtracted.

\(f(x)=x-7\) is a linear polynomial function. It has neither a denominator that could become zero nor a square root or logarithm that imposes a restriction. Hence, \(f(x)\) is defined for every real \(x\), so its domain is \(\mathbb{R}\). Option C is incorrect because \(f(7)=0\); there is no reason to exclude 7. Exam tip: Every polynomial function has real domain \(\mathbb{R}\).

The correct answer is option C: ℝ. The range is the collection of all possible outputs. Let y=7-2x, where x can be any real number. To see whether an arbitrary real y can occur, solve for x: 2x=7-y, so x=(7-y)/2. For every real y, this x is also real. Therefore every real output is possible, and the range is ℝ. Option A, [7,∞), is wrong because taking x=1 gives f(x)=5, which is below 7. Option B, (-∞,7], is wrong because taking x=-1 gives f(x)=9, which is above 7. Option C is correct: the non-constant linear expression has no upper or lower limit when x ranges over all real numbers. Option D, {7}, would be true only for a constant function whose output is always 7; here changing x changes the output. For example, x=0 gives 7, x=1 gives 5, and x=-1 gives 9. Exam cue: a linear function ax+b with a≠0 and domain ℝ has range ℝ.

The expression under a real square root must be non-negative. Thus, \(x-5\ge 0\), which gives \(x\ge 5\). Therefore, the domain is \([5,\infty)\). Option B is incorrect because \(x=5\) is also allowed, giving \(f(5)=0\). Exam tip: For a square-root function, set the radicand greater than or equal to zero to find the domain.

For \(\sqrt{x-5}\) to be real, we need \(x-5\ge 0\). Solving gives \(x\ge 5\). Hence the domain is \([5,\infty)\).

For a square-root function to have a real value, its radicand must be non-negative. Here the radicand is 8 − x, so we require 8 − x ≥ 0. Rearranging gives −x ≥ −8 and therefore x ≤ 8. Every real number less than 8 is allowed, and x = 8 is also allowed because √0 is defined. Thus the real domain is (−∞, 8], making option C correct. Option A reverses the inequality, option B incorrectly excludes the endpoint 8, and option D excludes 8 while allowing values greater than 8, which would make the radicand negative.

The denominator x − 6 must not be zero. It is zero when x = 6, so the function is undefined there and 6 is excluded from its domain.