Algebra of Functions:
Let us consider two functions, f: D1 →R and g: D2 →R .
We describe functions f + g , f – g , f.g and f/g as follows:
(i) f + g : D →R is a function defined by
(f + g)x = f(x)+g(x) where D = D1 ∩ D2
(ii) f – g : D → R is a function defined by
(f – g)x = f(x) –g(x) where D = D1 ∩ D2
(iii) f.g : D →R is a function defined by
(f. g)x = f(x). g(x) where D = D1 ∩ D2
(iv) f/g: D →R is a function defined by
(f/g)x =f(x)/g(x) where D = {x : x ∈ D1 ∩ D2, g(x) ≠ 0}
Illustration: Let $\large f(x) = \sqrt{6-x}$ , $\large g(x) = \sqrt{x-2}$ . Find f + g, f – g, f.g and f/g.
Solution: $\large (f+g)x = \sqrt{6-x} + \sqrt{x-2} $ , 2 ≤ x ≤ 6
$\large (f-g)x = \sqrt{6-x} – \sqrt{x-2} $ , 2 ≤ x ≤ 6
$\large (f.g)x = \sqrt{6-x} . \sqrt{x-2} = \sqrt{(6-x)(x-2)} $ , 2 ≤ x ≤ 6
$\large (f/g)x = \frac{\sqrt{6-x}}{\sqrt{x-2}} = \sqrt{\frac{6-x}{x-2}}$ , 2 < x ≤ 6
Exercise : Let f(x) = |sinx|, 0 ≤ x ≤ π and g(x) = |cosx|, -π/2 ≤ x ≤ π/2. Find f ± g , f.g and f/g and their respective domains .
Composite functions
Let f: X ∈ Y1 and g : Y1 ∈ Y be two given functions. Let a new function h(x) be derived in the following manner.
To obtain h(x) , we first take the f-image of an element x ∈ X so that f(x) ∈ Y1 , which is the domain of g(x).
Then take g-image of f(x) , i.e g(f(x)) which would be an element of Y. The adjacent figure clearly shows the steps to be taken.
The function ‘ h ‘ defined above is called the composition of f and g and is denoted by gof.
Thus (gof)x = g(f(x)).
Clearly Domain (gof) = {x : x ∈ Domain (f) , f(x) ∈ Domain(g)}
Similarly we can define, (fog)x = f(g(x)) and Domain (fog) = {x : x ∈ Domain (g), g(x) ∈ Domain (f)}.
In general fog ≠ gof.
Example : Two functions are defined as under,
$ \displaystyle f(x) = \left\{\begin{array}{ll} x + 1 \; , x \leq 1 \\ 2x+1 \; , 1 < x \leq 2 \end{array} \right. $
$ \displaystyle g(x) = \left\{\begin{array}{ll} x^2 \; , -1 \leq x < 2 \\ x + 2 \; , 2 \leq x \leq 3 \end{array} \right. $
Find fog & gof
Solution: $\large (fog)(x) = f(g(x)) = \left\{\begin{array}{ll} g(x) + 1 , g(x) \le 1 \\ 2g(x) + 1 , 1 < g(x) \le 2 \end{array} \right.$
Let us consider, g(x) ≤ 1:
(i) x2 ≤ 1, -1 ≤ x < 2 ⇒ -1 ≤ x ≤ 1 , -1 ≤ x < 2 ⇒ -1 ≤ x ≤ 1
(ii) x+2 ≤1, 2 ≤ x ≤ 3 ⇒ x ≤ -1 , 2 ≤ x ≤ 3 ⇒ x = φ
Let us consider, 1 < g(x) ≤ 2,
(iii) 1 < x2 ≤ 2, –1 ≤ x < 2 ⇒ x ∈ [ –√2, -1 ) ∪(1, √2], -1 ≤ x < 2
⇒ 1< x ≤ √2
(iv) 1< x+2 ≤ 2 , 2 ≤ x ≤ 3 ⇒ -1< x ≤ 0, 2 ≤ x ≤ 3, x =φ
$\large (gof)(x) = g(f(x)) = \left\{\begin{array}{ll} f^2(x) , -1 \le f(x) < 2 \\ f(x) + 2 , 2 \le f(x) \le 3 \end{array} \right.$
Let us consider –1 ≤ f(x) < 2:
(i) -1 ≤ x + 1< 2, x ≤ 1 ⇒ -2 ≤ x < 1, x ≤ 1 ⇒ -2 ≤ x < 1
(ii) -1 ≤ 2x + 1 < 2, 1< x ≤ 2 ⇒ -1 ≤ x< 1/2, 1< x ≤ 2 ⇒ x = φ
Let us consider 2 ≤ f(x) ≤ 3:
(iii) 2 ≤ x +1 ≤ 3, x ≤ 1 ⇒ 1 ≤ x ≤ 2 , x ≤ 1 ⇒ x = 1
(iv) 2 ≤ 2x + 1 ≤ 3, 1 < x ≤ 2 ⇒ 1 ≤ 2x ≤ 2, 1 < x ≤ 2
⇒ 1/2 ≤ x ≤ 1, 1 < x ≤ 2 ⇒ x = φ
$\large g(f(x)) = \left\{\begin{array}{ll} (x+1)^2 , -1 \le x < 1 \\ x + 3 , x=1 \end{array} \right.$
If we like we can also write g(f(x)) = (x+1)2, -2 ≤ x ≤ 1.
Exercise: If $\large f(x) = \left\{\begin{array}{ll} x^2 , x \le 0 \\ x , x > 1 \end{array} \right.$
and g(x) = – |x|, x ∈ R, then find fog and gof.