Real-Valued Functions, Their Domain and Range

The governing concept is the distinction between domain and range. The domain of a relation is the set of first coordinates, while its range is the set of second coordinates. From R = {(1,2),(2,3)}, the domain is {1,2} and the range is {2,3}. Comparing these sets, 1 occurs in the domain but not in the range. The element 2 belongs to both sets, and 3 belongs only to the range. Therefore option A is correct. Option B fails because 2 is common to domain and range, option C reverses the requested relationship because 3 is not in the domain, and option D is incorrect because the difference between the two sets is not empty; it is {1}.

The governing concept is the range of a relation. The range is the set of all second components, or output values, appearing in its ordered pairs. In R, the pairs are (2,5), (4,7), and (6,9). Their first components are 2, 4, and 6, so the domain is {2,4,6}. Their second components are 5, 7, and 9, so the range is {5,7,9}. Therefore option B is correct. Option A gives the domain, not the range. Option C mixes first and second components, while option D reverses the ordered pairs and therefore is not a set of range values. If a second component were repeated, it would still be listed only once because a set does not contain duplicate elements.

In the function \(f:A\to B\), \(A\) is the set of all input elements on which the function is defined, so it is called the domain. \(B\) is the codomain, while the set of actual output values is the range, which may be a subset of the codomain. In an exam, identify the first set in \(f:A\to B\) as the domain and the second as the codomain.

The governing concept is the range of a function. The range is the set of all actual output values, which are the second components of the ordered pairs. In the given function, the pairs are (1,5), (2,5), and (3,6). Hence the outputs are 5, 5, and 6. Because a set lists distinct elements only once, the repeated value 5 is written once, giving the range {5,6}. Option A is therefore correct. Option B, {1,2,3}, consists of the first components and is the domain, not the range. Option C incorrectly combines an input with the output values. Option D includes inputs 2 and 3, omits output 6, and therefore cannot be the range. The repeated output does not create a second distinct set element.

The domain of a function represented by ordered pairs is the set of their first components. Here, the first components are 2, 3, and 4, so the domain is \(\{2,3,4\}\). Option B lists the second components, which form the range. Exam tip: in an ordered pair \((x,y)\), the x-values form the domain and the y-values form the range.

Since \(2\) belongs to the given domain \(\{1,2,3\}\), substitute \(x=2\) in the rule: \(f(2)=2+2=4\). Option B incorrectly omits the added constant 2. Exam tip: to find a function value, replace \(x\) directly with the given input.

Apply the rule \(f(x)=x-1\) to every element of \(A\): \(f(2)=1\), \(f(3)=2\), and \(f(4)=3\). Hence, \(f(A)=\{1,2,3\}\). Option B is only the original set, not the set of images produced by the function. Exam tip: To find the image of a set, apply the function to each of its elements and collect the resulting values.

The set \(f(A)\) contains the function values of every element of \(A\). Here, \(f(1)=\sqrt{1}=1\), \(f(4)=\sqrt{4}=2\), and \(f(9)=\sqrt{9}=3\), so \(f(A)=\{1,2,3\}\). Option B is only the original set \(A\), not its image under the function. Exam tip: To find \(f(A)\), apply the function separately to each element of \(A\).

For \(f(x)=x^2+4\), every real value of \(x\) gives a defined output, so its domain is \(\mathbb{R}\). In option B, \(x=-2\) is excluded. Exam tip: for square-root functions, the expression inside the root must be \(\geq 0\).

The domain of a function is the set of all first components of its ordered pairs. Here, the first components are 5, 6, and 7, so the domain is \(\{5,6,7\}\). Option B is the range, not the domain, because it contains the second components 1 and 2. Exam tip: in an ordered pair \((x,y)\), collect the \(x\)-values for the domain and the \(y\)-values for the range.

Every element x \(\in A\) is mapped to 10, so the set of all output values is only \(\{10\}\). This is a constant function, and repeated output values are listed only once in the range. Exam tip: for a constant function \(f(x)=c\), the range is \(\{c\}\), provided the domain is non-empty.

To find f(A), apply the rule f(x) = x − 2 to every element of A: f(2) = 0, f(4) = 2, and f(6) = 4. Therefore, f(A) = {0, 2, 4}. Option B is incorrect because it is the original set before applying the function. Exam tip: To find the image of a set, apply the function rule separately to each element of the set.

Since \(f(a)=a^2\), we have \(a^2=16\), giving \(a=\pm4\). The question asks for the positive value, so \(a=4\) is correct. The value 16 is \(f(a)\), not \(a\). Exam tip: when taking a square root, consider both signs and then apply any stated sign condition.

The range of a function is the set of all actual output values. In both ordered pairs, the second component is 3, so the range is \(\{3\}\); a set lists 3 only once. \(\{3,4\}\) is the codomain, not the range, because no element of \(A\) maps to 4. Exam tip: To find the range from ordered pairs, collect their second components.

To find the range, substitute each domain value into the function: \(f(0)=1\), \(f(2)=3\), and \(f(4)=5\). Therefore, the range is \(\{1,3,5\}\). Option B merely repeats the domain, not the range. Exam tip: calculate the function value for every given input to obtain the range.

The polynomial \(x^2+1\) is defined for every real \(x\), so its domain is \(\mathbb{R}\). In \(\frac{1}{x-3}\), \(x=3\) is excluded. Exam tip: first check denominator and square-root restrictions.

The domain of a relation or function is the set of first components of its ordered pairs. Here, the first components are 3, 5, and 7, so the domain is {3, 5, 7}. Option B is the range, not the domain. In an exam, identify the domain by collecting the first element from every ordered pair.

The range of a function is the set of all second components of its ordered pairs. Here, the second components are 4, 6, 8, and 10, so the range is \(\{4,6,8,10\}\). Option B is the domain because it contains all the first components. Exam tip: in an ordered pair \((x,y)\), collect the y-values to find the range.

The set f(A) contains the function values corresponding to every element of A. Thus, f(4)=4−4=0, f(5)=5−4=1, and f(9)=9−4=5. Therefore, f(A)={0,1,5}. Option B is the original domain A, not its image under f. Exam tip: To find an image set, substitute each element of the domain into the function one by one.

For a function \(f:A\to B\), the image of every \(a\in A\) lies in \(B\). The set of all such images is the range, so the range is a subset of the codomain \(B\). It equals \(B\) only when the function is onto. Exam tip: the range is always contained in the codomain, but it need not fill the entire codomain.

Apply the function to each element of \(A\): \(f(-2)=|-2|=2\), \(f(0)=0\), and \(f(3)=3\). Hence, \(f(A)=\{2,0,3\}=\{0,2,3\}\). Option B is just the original set \(A\), not its image under \(f\). Exam tip: To find an image set, evaluate the function for every element of the given set and list each resulting value only once.

The linear function (f(x)=x+3) is defined for every real (x). In exams, the domain of a basic linear function is usually (\mathbb{R}).

The governing concept is the range of a real-valued square function. For every real number x, x² is non-negative, so f(x) ≥ 0. The least possible value is 0, and it is actually attained at x = 0 because f(0) = 0. Every positive real number is also attained: if y > 0, choose x = √y or x = −√y, and then x² = y. Thus the function produces 0 and every positive real value, but it can never produce a negative value. The range is therefore [0, ∞). Option B gives non-positive values, which is not the output of squaring; option C incorrectly includes negative numbers, and option D wrongly excludes the attained value 0.

Over the real numbers, \(\sqrt{x}\) is defined only when \(x\geq 0\). Therefore, zero and every positive real number belong to the domain, giving \([0,\infty)\). Option D excludes zero, but \(\sqrt{0}=0\) is defined. Exam tip: For a square-root function, require the expression inside the radical to be \(\geq 0\).

For \(f(x)=\frac{1}{x}\) to be defined, the denominator cannot be zero. So \(x\ne 0\). Therefore the domain is \(\{x\in\mathbb{R}\mid x\ne 0\}=( -\infty,0)\cup(0,\infty)\).