## Functions in Math:

There are many practical situations where we find a correspondence between two quantities. The correspondence may be shown in the form of a table and may be described by a formula connecting them or by a graph and may also be recorded as ordered pairs.

Mapping is a type of correspondence. It is a rule that assigns to the elements of one set, some elements of another set. The rule-following which the elements y of the set B may be obtained corresponding to the elements x of the set A is a relation between the two variables.

Let us consider the relation y = 2x -1. A few sets of values of x and y following this relation are given in the table below.

x | 1 | 2 | 3 | 4 | -2 |

y | 1 | 3 | 5 | 7 | -5 |

The corresponding mapping from set A, consisting of the values of x, to set B giving the values of y is shown in the figure below.

There may be several types of mappings. If each element of A corresponds to a single element in B, then the mapping is said to be a one-one mapping as shown below.

The set of values of x constitutes the ** domain** and the set of values of y forms the

**in the case of a mapping. Linear relations correspond to one-one mapping.**

*codomain*If two or more elements of the domain correspond to the same image in the codomain, then the mapping is said to be a many-one mapping. Relations of the type y = x^{2 }or y = x^{3} etc., correspond to many-one mappings as shown below.

If a single value of the domain corresponds to two or more values of the codomain, then the mapping is said to be one-many mapping. A relation of this type is x = y^{2} or y = ± √x as shown below.

Consider two non-empty sets A and B. A function f from the set A to the set B is a rule that assigns to each element x of A, one and only one element y of B. y, in this case, is called the image of x under f and is denoted by f(x). The domain of f is the set A and the codomain of f is the set B.

It is, thus, clear that all types of mappings denote some relation but only one-one or many-one mappings are functions. The definition of a function allows some elements of codomain not being the image of any element in the domain as shown in figures (i) and (ii). In the case of functions, set A must be exhausted. In figure (i), set B is the codomain of f but its range is the set {y_{1}, y_{2}, y_{3}, y_{4}}. Similarly in figure (ii), set B is the codomain of f and the set {y_{1}, y_{2}, y_{5}} is its range.

The ray diagram in the figure below shows a type of correspondence where the domain is not exhausted. This correspondence does not represent a function.

A function may, thus, be defined as a special type of relation which gives a set of ordered pairs of which no two pairs have the same first component. |

The ordered pairs (1, 1), (2, 3), (3, 5), (4, 7), (-1, -3), etc., denote a function as no two of the first entries have been repeated. The ordered pairs (1, 1), (1, -1), (4, 2), (4, -2), (9, 3), etc., do not correspond to a function, because in this case some of the first entries have been repeated.

Note that the set of all f-images of elements of A is known as the range of f and may be denoted by f(A).

Thus, f(A) = {f(x): x ∈ A} = Range of f. |

The Range of f = f(A) ⊆ B. In the figure below, A = domain, B = codomain and the set {y_{1}, y_{2}, y_{3}, y_{4}} is the range of f.

It is obvious that all functions are relations but all relations are not functions. The graph of a relation may make it clear whether the relation is a function or not. If a vertical line drawn across the graph of a relation intersects the graph at more than one point, then the relation is not a function.

All the graphs drawn in the figure below denote functions as a vertical line through each of the graphs intersecting the graph at a single point (it clearly indicates that no two first entries have been repeated).

The vertical line drawn through each of the graphs in the figure below intersects the graph at more than one point. None of the relations denotes a function.

### Independent and Dependent Variables:

If we denote a function as y = f(x), then x is known as the independent variable and y as the dependent variable.

### Evaluation of a Function:

We shall concentrate our attention mainly on the functions which are given in the form of an equation involving dependent and independent variables.

To evaluate the functional values, we generally isolate the dependent variable on the left side of the equation. For example, the equation 2x + 3y – 5 = 0 is generally written as

y = (5 – 2x)/3 or f(x) = (5 – 2x)/3 The value of f(x) when x = 4 is f(4) = (5 – 2 x 4)/3 = -1 |

### Examples:

Given f: R Example-→ R defined as f(x) = 4x -3, find (i) f(4) (ii) f(-2) (iii) f(1/2) f: R Solution-→ R and f(x) = 4x -3(i) f(4) = 4(4) – 3 = 13 (ii) f(-2) = 4(-2) – 3 = -8 – 3 = -11 (iii) f(1/2) = 4(1/2) – 3 = 2 – 3 = -1 |

A function f is defined on the set R as f(x) = 3xExample-^{2} – 1, for all x ∈ R. Compute (i) the image of 2 (ii) the pre-image corresponding to image 74 (iii) the value of f at -√3 f(x) = 3xSolution-^{2} – 1(i) f(2) = 3(2) ^{2} – 1 = 11∴ image of 2 is 11. (ii) Let x be the pre-image corresponding to image 74. ∴ f(x) = 74 ⇒ 3x ^{2} – 1 = 74⇒ 3x ^{2} = 75⇒ x ^{2} = 25⇒ x = ±5 (iii) f(-√3) = 3(-√3) ^{2} – 1 = 9 – 1 = 8 |

Find the domains of the following functions:Example-(i) f(x) = √(2 + x – x^{2}) f(x) is defined ifSolution-2 + x – x ^{2} ≥ 0⇒ x ^{2} – x – 2 ≤ 0⇒ x ^{2} – 2x + x – 2 ≤ 0⇒ x (x – 2) + 1 (x – 2) ≤ 0 ⇒ (x +1) (x – 2) ≤ 0 ⇒ -1 ≤ x ≤ 2 ∴ D _{f} = [-1, 2](ii) f(x) = x^{2} + 3 f(x) is defined for all values of x ∈ R.Solution-∴ D _{f} = R(iii) f(x) = 5/(2 – x^{2}) f(x) is defined ifSolution-2 – x ^{2} ≠ 0⇒ 2 ≠ x ^{2}⇒ x ^{2} ≠ 2⇒ x ≠ ±√2 ∴ D _{f} = R – {√2, -√2}(iv) f(x) = 2 / | x | f(x) is defined ifSolution-| x | ≠ 0 ⇒ x ≠ 0 ∴ D _{f} = R – {0}(v) f(x) = x/(x^{2} – 3x + 2) f(x) is defined ifSolution-x ^{2} – 3x + 2 ≠ 0⇒ x ^{2} – 2x – x + 2 ≠ 0⇒ x (x – 2) – 1 (x – 2) ≠ 0 ⇒ (x – 1) (x – 2) ≠ 0 ⇒ x ≠ 1, x ≠ 2 ∴ D _{f} = R – {1, 2}(v) f(x) = √(x – 1) / (x^{2} – 4) x -1 ≥ 0Solution-⇒ x ≥ 1 ……….(i) and x ^{2} – 4 ≠ 0⇒ x ^{2} ≠ 4⇒ x ^{2} ≠ ±2 ……….(ii)∴ D _{f} = (x ≥ 1) ∩ x ≠ ±2 = (1, 2) ∪ (2, ∞)(vii) f(x) = √[(x – 2)/(x + 3)] f(x) is defined ifSolution-x – 2 ≥ 0 and x + 3 > 0 ⇒ x ≥ 2 and x > -3 ∴ The domain of f is x ≥ 2 |

Find the range of each of the following functions.Example-(i) f(x) = 1 / | x | f(x) = 1 / | x |Solution-| x | > 0 ∀ values of x ∴ The range of f is R ^{+}(ii) f(x) = x^{2}/(1 + x^{2}) Let y = xSolution-^{2}/(1 + x^{2})⇒ y + x ^{2}y = x^{2}⇒ y = x ^{2} – x^{2}y⇒ y = x ^{2} (1 – y)⇒ x ^{2} = y/(1 – y)⇒ x = √[y/(1 – y)] For real values of n, y/(1 – y) ≥ 0 ⇒ y/[- (y – 1)] ≥ 0 ⇒ y/(y – 1) ≤ 0 ⇒ 0 ≤ y ≤ 1 ∴ Range = (0, 1) (iii) 1/(2 – cos 3x) Let f(x) = 1/(2 – cos 3x)Solution--1 ≤ cos 3x ≤ 1 ⇒ 1 ≥ – cos 3x ≥ -1 ⇒ 2 + 1 ≥ 2 – cos 3x ≥ 2 – 1 ⇒ 3 ≥ 2 – cos 3x ≥ 1 ⇒ 1/3 ≤ 1/(2 – cos 3x) ≤ 1 ⇒ 1/3 ≤ f(x) ≤ 1 ∴ Range = (1/3, 1) (iv) f(x) = 1 – | x | f(x) = 1 – | x |Solution-⇒ | x | ≥ 0 ⇒ – | x | ≤ 0 ⇒ 1 – | x | ≤ 0 + 1 ⇒ f(x) ≤ 1 ∴ Range is (-∞, 1) |

If f(x) = (1/2) (aExample- ^{x} + a^{-x}) and g(x) = (1/2) (a^{x} – a^{-x}), show that f(x + y) = f(x) f(y) + g(x) g(y). f(x) = (1/2) (aSolution-^{x} + a^{-x}) and g(x) = (1/2) (a ^{x} – a^{-x})Now, f(x + y) = (1/2) [a ^{x+y} + a^{-(x+y)}]⇒ f(x + y) = (1/2) [a ^{x}a^{y}+ a^{-x}a^{-y}] ……….(i)Now, f(x) f(y) + g(x) g(y) = (1/2) (a ^{x} + a^{-x}) (1/2) (a^{y} + a^{-y}) + (1/2) (a^{x} – a^{-x}) (1/2) (a^{y} – a^{-y})⇒ f(x) f(y) + g(x) g(y) = (1/4) [a ^{x}a^{y} + a^{x}a^{-y} + a^{-x}a^{y} + a^{-x}a^{-y}] + (1/4) [a^{x}a^{y} – a^{x}a^{-y} – a^{-x}a^{y} + a^{-x}a^{-y}]⇒ f(x) f(y) + g(x) g(y) = (1/4) [a ^{x}a^{y} + a^{x}a^{-y} + a^{-x}a^{y} + a^{-x}a^{-y} + a^{x}a^{y} – a^{x}a^{-y} – a^{-x}a^{y} + a^{-x}a^{-y}]⇒ f(x) f(y) + g(x) g(y) = (1/4) [2 a ^{x}a^{y} + 2 a^{-x}a^{-y}]⇒ f(x) f(y) + g(x) g(y) = (1/2) [a ^{x}a^{y}+ a^{-x}a^{-y}] ……….(ii)From (i) and (ii):- f(x + y) = f(x) f(y) + g(x) g(y) |

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