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) = x3 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) = x3 + 3x + 1.
- (f – g) (x) = f(x) – g(x) = x3 – (3x + 1) = x3 – 3x – 1.
- ($\alpha$f) (x) = $\alpha$ f(x) = 6x3
- ($\alpha$g) (x) = $\alpha$ g(x) = 6 (3x + 1) = 18x + 6.
- (fg) (x) = f(x) g(x) = x3 (3x +1) = 3x4 + x3.
- ($\frac{f}{g}$) (x) = $\frac{f(x)}{g(x)}$ = $\frac{x^3}{3x+1}$, provided x ≠ –$\frac{1}{2}$.
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