# Algebra of Real Functions

Let f : X → R and g : X → R be any two real functions, where X $\subset$ R.

**(i) Addition of two real functions**

For adding two real functions let us define the functions f and g such that f: X ⟶**R** and g: X ⟶**R **are two real functions such that X is a subset of R.

Then (f + g): X ⟶**R **can be defined as:

(*f* + g)(x) = *f*(x) + g(x), *for all x ϵ X*

**(ii) Subtraction of two real functions **

For subtracting two real functions let us define the functions f and g such that f: X ⟶**R** and g: X ⟶**R **are two real functions such that X is a subset of R.

Then (f – g): X ⟶**R** can be defined as:

(*f* + g)(x) = *f*(x) + g(x), *for all x ϵ X*

**(iii) Multiplication by a scalar**

Let us define a real function f such that f: X ⟶**R, **X⊆**R** and a $\alpha$ be a real scalar quantity. Then the product of scalar $\alpha$ and the function f is also a function defined from X to **R **as:

($\alpha$ *f*)(x) = $\alpha$*f* (x), *for all x ϵ X*

**(iv) Multiplication of two real functions**

For multiplying two real functions let us define the functions f and g such that f: X ⟶**R** and g: X ⟶**R **are two real functions such that X is a subset of R.

Then fg: X ⟶**R** can be defined as:

*f*(g(x)) = *f*(x) g(x), *for all x ϵ X*

**(v) Quotient of two real functions**

For determining the quotient of two real functions let us define the functions f and g such that f: X ⟶**R** and g: X ⟶**R **are two real functions such that X is a subset of R.

Then f/g: X ⟶**R** can be defined as:

$\left(\frac{f}{g}\right)$(x) = $\frac{f(x)}{g(x)}$, g(x) $\neq$ 0

Given that g (x) ≠ 0, for all x *ϵ* X

**Simple Problem: **

Let f(x) = x^{3 }and g(x) = 3x + 1 and a scalar, $\alpha$ = 6. Find

- (f + g) (x)
- (f – g) (x)
- ($\alpha$f) (x)
- ($\alpha$g) (x)
- (fg) (x)
- ($\frac{f}{g}$) (x)

**Answer :**

We have,

- (f + g) (x) = f(x) + g(x) = x
^{3 }+ 3x + 1. - (f – g) (x) = f(x) – g(x) = x
^{3 }– (3x + 1) = x^{3 }– 3x – 1. - ($\alpha$f) (x) = $\alpha$ f(x) = 6x
^{3 } - ($\alpha$g) (x) = $\alpha$ g(x) = 6 (3x + 1) = 18x + 6.
- (fg) (x) = f(x) g(x) = x
^{3 }(3x +1) = 3x^{4}+ x^{3}. - ($\frac{f}{g}$) (x) = $\frac{f(x)}{g(x)}$ = $\frac{x^3}{3x+1}$, provided x ≠ –$\frac{1}{2}$.

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