Function of One Variable

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

1. Set A is called function
2. Set B is called co-domain
3. y is called image of x under f  i.e. y = f(x)
4. x is called pre-image of y
5. 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 = x²

Solution:

Given y = x²

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 = x²  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:

\(y = \sqrt{9 – x^2}\) \(\Rightarrow y ^ 2 = 9 – x ^ 2\) \(\Rightarrow x = \sqrt{9 – y^2}\)

 

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) = x³ + 2

Solution:

Since y = x³  + 2

or   x³ = 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^-1 and the graph of f^-1 is obtained by reflecting the graph of f about line y = x