Function of One Variable

By Suresh Chand

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  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

Important Questions