**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) = x^{2 }+ 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 (x^{3}) + 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].