Functions as a special kind of relation

Each element (1) and (2) of (A) has exactly one image in (B). In exams, first count the image of every domain element to test a function.

A function is represented by ordered pairs (x, f(x)), where the first component is an element of the domain and the second component is its uniquely assigned image. The rule here is f(x) = x². Substituting each domain element gives f(1) = 1² = 1, f(2) = 2² = 4, and f(3) = 3² = 9. Therefore the graph or ordered-pair representation is {(1,1),(2,4),(3,9)}, so option A is correct. Option C reverses the input-output direction, treating image values as first components. Option B uses values that do not follow x², and option D reverses or rearranges the outputs without applying the stated rule. Every domain element appears exactly once, confirming that this is a function.

In the notation f: A → B, the set A contains the allowed input values and is called the domain of the function. The set B is the codomain, while the range consists only of the actual output values produced by elements of A. An inverse is a separate function or relation, not the name of A. Therefore option A is correct.

This is a piecewise-defined function, so the value must be found by checking which condition contains the given input. For x=2, the first condition x>2 is false because 2 is not greater than 2. The second condition is x≤2, and it is true because equality is included: 2≤2. Therefore the second rule applies and h(2)=0. Option B is correct. Option A would be obtained only if the condition were x≥2 or if the first branch included equality. Option C is impossible because 2 is not an element of the codomain {0,1}, and option D is incorrect because the second branch explicitly defines the value at x=2. Careful attention to ≤ is the key concept.

In function notation f:A→B, an input a from the domain A is assigned an output b in the codomain B. The statement f(a)=b specifically says that b is the image of a under f. Thus option A is correct. The input a is called the preimage, or inverse image, of b when discussing this particular correspondence, while b is its image. The domain is the entire set A, not an individual output, so option B is incorrect. A relation is a general set of ordered pairs and is not the name of b itself, so option D is also incorrect. The notation is read as “the image of a under f is b,” which directly establishes the required terminology.

The governing idea is that an ordered pair (x,y) belongs to the function precisely when y=f(x). The domain is A={0,1,2}, and the rule is f(x)=x+2. Substitution gives f(0)=0+2=2, f(1)=1+2=3, and f(2)=2+2=4. Thus the function contains exactly the pairs (0,2), (1,3), and (2,4). The proposed pair (2,3) does not belong because applying the rule to x=2 produces 4, not 3. Options A, B, and C each satisfy y=x+2 and are therefore valid members. Hence option D is correct. The codomain includes all these outputs, so no additional restriction changes the result.

A function must assign exactly one output to every input in its domain. For every real number x, |x| is a uniquely determined non-negative real number, so f: ℝ → ℝ is indeed a function. It is not required that different inputs have different outputs; that stronger property is injectivity. Since f(2) = f(−2) only shows that the function is not one-one, option A is correct.

To be a function from A to B, every element of the domain A must appear exactly once as a first component, and its second component must be an element of B. Option A has first components 1, 2, and 3, each appearing once, with images 4, 5, and 6; all three images belong to B. Thus it satisfies the function definition. In option B, input 1 is paired with both 4 and 5, while input 3 is missing, so it fails. Option C omits domain element 3, and option D omits domain element 1. A function need not use every element of B, but it must assign one image to every element of A. Therefore A is the only correct answer.

The range of a function is the set of all actual output values, or images, produced by elements of the domain. In an ordered pair (x, y), the first component x is an input and the second component y is its image. For f = {(1, u), (2, v), (3, w)}, the second components are u, v, and w. Therefore, Range(f) = {u, v, w}. Option A is the domain, not the range. Option C leaves out w, even though w is produced by 3, and option D includes only one of the three output values. Hence option B is correct. In this example, the range happens to equal the codomain B because every element of B is attained.

A function from A to B must assign exactly one image in B to every member of A. Since A = {1, 2, 3}, the first components must include 1, 2, and 3, with each appearing once. Option A contains pairs for 1 and 2 only; element 3 has no image. Thus option A is not a function from A to B. Option B assigns one image to each domain element, and option D does the same. Option C is also a function even though several inputs have the same image, because a function may be many-to-one. Therefore, the missing-domain-element condition makes A the unique correct answer.

The defining condition for a function from A to B is that every element of A must have exactly one image in B. If one element of A points to two distinct elements of B, that input has two images, so the relation is not single-valued. This single violation is sufficient to conclude that the relation is not a function, regardless of the number of elements in B or whether A and B are equal. Therefore option A is correct. Option B incorrectly reverses the function condition. Options C and D introduce irrelevant restrictions: a codomain may contain any appropriate number of elements, and a function does not require A and B to be equal. Other arrows in the diagram cannot repair the fact that one input has two outputs.

The governing concept is evaluation of a piecewise-defined function at a boundary point. The input is x = −1. The first rule applies for x ≤ −1, whereas the second rule applies only for x > −1. Since −1 satisfies the first condition, we must use f(x) = x² + 1. Therefore f(−1) = (−1)² + 1 = 1 + 1 = 2. Although the second formula would produce 2 − (−1) = 3, that formula is not permitted at x = −1 because its condition is strict. A function assigns one value to each input in its domain, so the answer is neither two values nor undefined. Thus option A is correct.

To find \(f(0)\), substitute 0 for x in the rule: \(f(0)=4-0=4\). Therefore, the answer is 4.

Substituting \(x=11\) into the function gives \(f(11)=\frac{11-3}{4}=\frac{8}{4}=2\). Therefore, option C is correct. Option B results from an incorrect simplification; the numerator must first be calculated as \(11-3=8\). Exam tip: To find the value of a function, replace \(x\) with the given input and then simplify carefully.

Substitute x = 4 throughout the function: f(4) = 4² − 2(4) = 16 − 8 = 8. Therefore, the correct answer is 8.