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Continue Shopping ```Algebra of Functions Given functions f : D → R and g : D → R, we describe functions f + g, f - g, gf an f/g as follows:

f + g : D → R is a function defined by (f + g) (x) = f(x) + g(x)

f - g : D → R is a function defined by (f - g)(x) = f(x) - g(x)

fg : D → R is a function defined by (fg) (x) = f(x) g(x)

f/g : C → R is a function defined by (f/g) (x) = f(x)/g(x) , g(x) ≠ 0,

where C = {x ε D : g(x) ≠ 0}.

Illustration: Let f : [-1, 1] → R and g : R → R be functions defined by f(x) = √(1-x2) and g(x) = x3 + 1. Find the domain of f + g, fg and f/g.

Solution:

(f+g)(x) = f(x)+g(x) = √(1-x2) + x3 + 1, (f-g)(x) - g(x) = √(1-x2) - x3 = 1,

(fg)(x) = f(x) g(x) = (x3 + 1) √(1-x2)  and  (f/g)(x) = f(x)/g(x)  = (√(1-x2))/(x3- 1), x ≠ 1.

The domain of each of f + g. f - g and fg is [-1, 1] and of f/g is

{x ε [-1, 1] : g(x) ≠ 0} = (-1, 1].

Illustration: Let f(x) = Describe the function f/g.

Solution:

We redefine the functions f(x) and g(x) in the intervals as shown below: At askIITians we provide you free study material on Set Theory and Functions and you get all the professional help needed to get through IIT JEE and AIEEE easily. AskIITians also provides live online IIT JEE preparation and coaching where you can attend our live online classes from your home!

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