Function of One Variable

 

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 $f:A\to B$ 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) = $\left\{f\right(x):\forall x\epsilon A\}$

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 $y\ge 0$, x is defined. Thus, y is set of all non-negative and real number. Hence, range is $y\ge 0$ 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 $9-x^2\ge 0$

i.e. $x^2–9\ge 0$

i.e. $(x-3)(x+3)\le 0$

i.e. $-3\le x\le 3$

Hence, domain is $-3\le x\le 3$ 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 $9-y^2\ge 0$

i.e. $y^2–9\ge 0$

i.e. $(y-3)(y+3)\le 0$

i.e. $-3\le y\le 3$

But given $y=\sqrt{9-x^2}$, which is non-negative. hence range is $0\le y\le 3$ 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

$f\left(x\right)=\frac{1}{x^{-4}}$

Solution:

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

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

Therefore, the given $f\left(x\right)=x^{-4}$ 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) = $\frac{f\left(x\right)}{g\left(x\right)}$

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 $f\left(x\right)=\sqrt{x}$ and [M]$g\left(x\right)=\sqrt{2-x}$

Find the each function and its domain

a) fog   b) gof    c) fof     d) gog

Solution:

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

$\sqrt{\sqrt{2-x}}=(2-x{)}^{\frac{1}{4}}$

To find domain of fog

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

Domain of g(x) is

$2–x\ge 0$ $x\le 2$

Here, (-∞, 2] is domain of fog.

 

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

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

Domain of gof = [0, 4]

 

(c)         $f.f\left(x\right)=f\left(f\right(x\left)\right)=f\left(\sqrt{x}\right)$

$=\sqrt{\sqrt{x}}=x^{\frac{1}{4}}$

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

Domain of fof = [0, ∞]

 

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

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

$f^{-1}\left(y\right)=x\iff f\left(x\right)=y$

 

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    $f^{-1}\left(y\right)=(y–2{)}^{\frac{1}{3}}$

or    $f^{-1}\left(x\right)=(x–2{)}^{\frac{1}{3}}$ 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