# Intervals

Closed Interval: Let a and b be two given real numbers such that a < b. Then the set of all real numbers x such that a ≤ x ≤ b is called a closed interval and is denoted by [a, b] i.e., [a, b] = {x ϵ R| a ≤ x ≤ b}

Open Interval: Let a and b be two given real numbers such that a < b. Then the set of all real numbers x such that a ≤ x ≤ b is called a closed interval and is denoted by (a, b) i.e., (a, b) = {x ϵ R| a ≤ x ≤ b}

Semi-Closed OR Semi Open Interval: If a, b are two given real numbers such that a < b then the sets (a, b) = {x ϵ R| a ≤ x ≤ b} and (a, b) = {x ϵ R| a ≤ x < b} are known as semi-closed or semi-open intervals and are also denoted by ] a, b [ and [a, b] respectively.

REAL FUNCTIONS:

Real Function: If the domain and co-domain of a function are subsets of R (set of all real numbers). It is called a real valued function or in short a real function.

Description of Areal Function: If f is a real valued function with finite domain, then f can be described by listing the values which it attains at different points of its domain. However, if the domain of a real function is an infinite set, then, f cannot be described by listing the values at points in its domain. In such cases real functions are generally described by some general formula or rule like f (x) = x2 + 1 or f (x) = 2sinx + 3 etc.

Example: If $$f\left( x \right)=x+\frac{1}{x}$$, prove that $${{\left[ f\left( x \right) \right]}^{3}}=f\left( {{x}^{3}} \right)+3f\left( \frac{1}{x} \right)$$.

Solution: We have,

$$f\left( x \right)=x+\frac{1}{x}$$.

$$f\left( {{x}^{3}} \right)={{x}^{3}}+\frac{1}{{{x}^{3}}}$$.

Now,

$${{\left[ f\left( x \right) \right]}^{3}}={{\left( x+\frac{1}{x} \right)}^{3}}$$.

$$=\left( {{x}^{3}}+\frac{1}{{{x}^{3}}} \right)+3\left( x+\frac{1}{x} \right)$$.

= f (x3) + 3 f (x).

$$=f\left( {{x}^{3}} \right)+3f\left( \frac{1}{x} \right)\,$$.

DOMAIN AND RANGE OF A REAL FUNCTION:

Domain: Generally real functions in calculus are described by some formula and their domains are not explicitly stated. In such cases to find the domain of a function f (say) we use the fact that the domain is the set of all real numbers x for which f (x) is a real number.

Range: As discussed, the range of a function f (x) is the set of values of f (x) which it attains at points in its domain. For a real function the co-domain is always a subset of R. So, range of a real function f is the set of all points  such that y = f (x) where x ϵ Dom f (x)

ALGORITHM:

Step 1: Put f (x) = y

Step 2: Solve the equations in step 1 for x to obtain x = ϕ (y).

Step 3: Find the values of y for which the values of x, obtained from x = ϕ (y) are in the domain of f.

Step 4: The set of values of y obtained in step 3 is the range of f.

Example: Find the domain and range of the function \[f\left( x \right)=\frac{1}{2-\cos 3x}[/latex].

Solution: We have, \[f\left( x \right)=\frac{1}{2-\cos 3x}[/latex]

Domain: We have,

-1 ≤ cos 3x ≤ 1 for all x ϵ R,

⇒ -1 ≤ -cos3x ≤ 1

⇒ -1 ≤ 2 – cos 3x ≤ 3 for all x ϵ R,

⇒ f (x) is defined for all x ϵ R

So, domain (f) = R.

Range: Let f (x) = y Then

⇒ $$\frac{1}{2-\cos 3x}=y$$.

⇒ $$2-\cos 3x=\frac{1}{y}$$.

⇒ $$\cos 3x=2-\frac{1}{y}$$.

⇒ $$-1\le 2-\frac{1}{y}\le 1\,$$.

⇒ $$-3\le -\frac{1}{y}\le -1\,$$.

⇒ 3 ≥ 1/y ≥ 1

⇒ ⅓ ≤ y ≤ 1

⇒ y ϵ [⅓, 1]

So, range of f = [⅓, 1].