The goal of study function of one variable is after understanding functions. In this unit functions can be represent by different ways.

### Function of One Variable

A function of one variable x is a prescription y(x), which calculates a number, the function value, for any feasible value of the variable x.

Let A and B be the non-empty sets. then a rule A to B is called function if every element of domain A has unique association with elements of co-domain B.

Let \(f:A \rightarrow B \) be a function

- Set A is called function
- Set B is called co-domain
*y*is called image of x under*f*i.e.*y =**f*(x)*x*is called pre-image of*y*- Range oof
*f*is denoted by*f(A)*and defined by*f(A) =*\(\left \{ f(x) : \forall \enspace x \enspace \varepsilon \enspace A \enspace \right \}\)

## Identify Domain and Range

### i) *y = x*^{2}

^{2}

Solution:

Given *y = x ^{2}*

**For domain**, for all real values of x, y exists. So, domain is set of all real number i.e. domain is (-∞, ∞)

**For Range**, *y = x ^{2} * or \(x = \sqrt{y}\)

*.*For all \(y \geq 0\), x is defined. Thus, y is set of all non-negative and real number. Hence, range is \(y \geq 0\) or [0, ∞).

### ii) \(y = \frac{1}{x}\)

Solution:

**For Domain, **when x = 0, y doesn’t exist, it means y is exists for all real number except 0. The domain is set of all real number except 0 i.e. (-∞, 0) U (0, ∞)

**For Range, **\(y = \frac{1}{x}\) or \(x = \frac{1}{y}\). Here x is exists for all real number except y = 0. Thus, the range is set of all real number except o i.e. (-∞, 0) U (0, ∞)

### iii) \(y = \sqrt{9-x^{2}}\)

Solution:

**For Domain:**

y is exists for \(9-x^{2} \geq 0\)

i.e. \(x^{2} – 9 \geq 0\)

i.e. \((x-3)(x+3) \leq 0\)

i.e. \(-3 \leq x \leq 3\)

Hence, domain is \(-3 \leq x \leq 3\) i.e. [-3, 3]

**For Range:**

Here, x exists for \(9-y^{2} \geq 0\)

i.e. \(y^{2} – 9 \geq 0\)

i.e. \((y-3)(y+3) \leq 0\)

i.e. \(-3 \leq y \leq 3\)

But given \(y = \sqrt{9-x^{2}}\), which is non-negative. hence range is \(0 \leq y\leq 3\) i.e. [0, 3]

## Even and Odd Function: Symmetry

A function y = f(x) is even function if f(-x) = f(x), and odd function if f(-x) = -f(x) for every value of x.

Note that the graph of even function is symmetrical about y-axis and the graph of odd function is symmetrical about origin.

### Identify the Symmetry

#### \(f(x) = \frac{1}{x^{-4}}\)

Solution:

Since, \(f(x) = x^{-4} = \frac{1}{x^{4}}\)

So, \(f(-x) = \frac{1}{(-x)^{4}} = \frac{1}{x^4} = f(x)\)

Therefore, the given \(f(x) = x^{-4}\) is even function.

## Combination of Functions

Like numbers function can be added, subtracted, multiplied and divided to product new function. Let f an g are two functions then f + g, f – g, f.g and \(\frac{f}{g}\) are new function which are defined by

Sum: (f + g)(x) = f(x) + g(x)

Difference: (f – g)(x) = f(x) – g(x)

Product: (f.g)(x) = f(x) . g(x)

Quotient: (\(\frac{f}{g}\))(x) = \(\frac{f(x)}{g(x)}\)

### Composite Functions

Let f and g are functions, then composite function fog is defined by

*(fog)(x) = f(g(x))*

Note that the domain of *fog(x)* is intersection of domain of g(x) and f(g(x)).

**Example: **If \(f(x) = \sqrt{x}\) and [M]\(g(x) = \sqrt{2-x}\)

Find the each function and its domain

a) fog b) gof c) fof d) gog

**Solution:**

(a) \(fog(x) = f(g(x)) = f(\sqrt{2-x})\)

\(\sqrt{\sqrt{2-x}} = (2-x)^{\frac{1}{4}}\)To find domain of fog

Fist we find domain of \(g(x) = \sqrt{2-x}\)

Domain of g(x) is

\(2 – x \geq 0\) \(x \leq 2\)Here, (-∞, 2] is domain of fog.

(b) \(gof(x) = f(f(x)) = f(\sqrt{x}) = \sqrt{2 – \sqrt{x}}\)

To Find the domain of gof, Follow steps of question (a)

Domain of gof = [0, 4]

(c) \(f.f(x) = f(f(x)) = f(\sqrt{x})\)

\(= \sqrt{\sqrt{x}} = x^{\frac{1}{4}}\)To Find the domain of fof, Follow steps of question (a)

Domain of fof = [0, ∞]

(d) \(g.g(x) = g(g(x)) = g(\sqrt{2 – x}) = \sqrt{2 – \sqrt{2 – x}}\)

To Find the domain of gog, Follow steps of question (a)

Domain of fof = [-2, 2]

## Inverse of Function

Let *f* be a one to one function with domain A and range B. Then its inverse function *f ^{-1}* has domain B and range A and is defined by

\(f^{-1}(y) = x \Leftrightarrow f(x) = y\)

**Example: Find inverse of f(x) = x ^{3} + 2**

**Solution:**

Since y = x^{3} + 2

or x^{3} = y – 2

or \(x = (y – 2)^{\frac{1}{3}}\)

or \(f^{-1}(y) = (y – 2)^{\frac{1}{3}}\)

or \(f^{-1}(x) = (x – 2)^{\frac{1}{3}}\) is the required formula for inverse function.

The domain of *f* is range of *f ^{-1}* and range of domain of

*f*and the graph of

^{-1}*f*is obtained by reflecting the graph of

^{-1}*f*about line y = x