A function imposes certain limitations by virtue of definition itself. We have seen such restriction with respect to radical functions in which polynomial inside square root needs to be non-negative. We have also seen that denominator of a rational function should not be zero. We shall learn about different functions in subsequent modules. Here, we consider few examples for illustration :

1 : f x = log a 3 x 2 − x + 4 f x = log a 3 x 2 − x + 4

Here, logrithmic function is defined for a ∈ 0,1 ∪ 1, ∞ a ∈ 0,1 ∪ 1, ∞ and

3 x 2 − x + 4 > 0 3 x 2 − x + 4 > 0

2 : f x = arcsin 3 x 2 − x + 4 f x = arcsin 3 x 2 − x + 4

Here, arcsine function is defined in the domain [-1,1]. Hence,

- 1 ≤ 3 x 2 − x + 4 ≤ 1 - 1 ≤ 3 x 2 − x + 4 ≤ 1

It is clear that we need to have clear understanding of algebraic inequalities as function definitions are defined with certain condition(s).

Function inequality compares function to zero. There are four forms :

1 : f(x) < 0

2 : f(x) ≤ 0

3 : f(x) > 0

4 : f(x) ≥ 0

Here, f(x) < 0 and f(x) >0 are strict inequalities as they confirm the notion of “less than” and “greater than”. There is no possibility of equality. If a strict inequality is true, then non-strict equality is also true i.e.

1 : If f(x) > 0 then f(x) ≥ 0 is true.

2 : If f(x) ≥ 0 then f(x) > 0 is not true.

3 : If f(x) < 0 then f(x) ≤ 0 is true.

4 : If f(x) ≤ 0 then f(x) < 0 is not true.

Further, we may be presented with inequality which compares function to non-zero value :

3 x 2 − x ≤ - 4 3 x 2 − x ≤ - 4

However, such alterations are equivalent expressions. We can always change this to standard form which compares function with zero :

3 x 2 − x + 4 > 0 3 x 2 − x + 4 > 0

Some important definitions/ results are enumerated here :

Equivalently, we may state above deductions symbolically.

If x > y , then : If x > y , then :

x + a > y + a x + a > y + a

a x > a y ; a > 0 a x > a y ; a > 0

a x < a y ; a < 0 a x < a y ; a < 0

x 2 > y 2 ; x , y > 0 x 2 > y 2 ; x , y > 0

1 x < 1 y ; when “x” and “y” have same sign. 1 x < 1 y ; when “x” and “y” have same sign.

It is evident that we can deduce similar conclusions with the remaining three inequality signs.

Linear function is a polynomial of degree 1. A linear inequality can be solved for intervals of valid “x” and “y” values, applying properties of inequality of addition, subtraction, multiplication and division. For illustration, we consider a logarithmic function, whose argument is a linear function in x.

f x = log e 3 x + 4 f x = log e 3 x + 4

The argument of logarithmic function is a positive number. Hence,

⇒ x > - 4 3 ⇒ x > - 4 3

Therefore, interval of x i.e. domain of logarithmic function is - 4 / 3, ∞ - 4 / 3, ∞ . The figure shows the values of “x” on a real number line as superimposed on x-axis. Note x= - 4/3 is excluded.

Figure 1: Domain is traced on x-axis.

Graph of logarithmic function

When f(x) = 0,

3 x + 4 = e f x = e 0 = 1 3 x + 4 = e f x = e 0 = 1

⇒ x = - 1 ⇒ x = - 1

It means graph intersects x-axis at x=-1 as shown in the figure. From the figure, it is clear that range of function is real number set R.

Example 1

Problem : A linear function is defined as f(x)=2x+2. Find valid intervals of “x” for each of four inequalities viz f(x)<0, f(x) ≤ 0, f(x) > 0 and f(x) ≥ 0.

Solution : Here, given function is a linear function. At y=0,

f x = 2 x + 2 = 0 f x = 2 x + 2 = 0

⇒ x = - 1 ⇒ x = - 1

At x=0,

f x = 2 f x = 2

We draw a line passing through these two points as shown in the figure. From the figure, we conclude that :

Figure 2: Graph is continuous for all values of x.

Graph of linear function

⇒ f x < 0 ; x ∈ - ∞ , - 1 ⇒ f x < 0 ; x ∈ - ∞ , - 1

⇒ f x ≤ 0 ; x ∈ ( - ∞ , - 1 ] ⇒ f x ≤ 0 ; x ∈ ( - ∞ , - 1 ]

⇒ f x > 0 ; x ∈ - 1, ∞ ⇒ f x > 0 ; x ∈ - 1, ∞

⇒ f x ≥ 0 ; x ∈ [ - 1, ∞ ) ⇒ f x ≥ 0 ; x ∈ [ - 1, ∞ )