Gilbert

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answers: 1

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Answer:

The functions we have are f(x) and g(x)`!=0` .

(f+g)(x) = f(x) + g(x)

(f-g)(x) = f(x) - g(x)

(f*g)(x) = f(x)*g(x)

(f/g)(x) = f(x)/g(x)

As g(x)`!=0`, f(x)/g(x) is always defined.

As an example take the functions f(x) = 3x + 2 and g(x) = x^2 + 1

(f+g)(x) = f(x) + g(x) = 3x + 2 + x^2 + 1 = x^2 + 3x + 3

(f-g)(x) = f(x) - g(x) = 3x + 2 - x^2 - 1 = -x^2 + 3x + 1

(f*g)(x) = f(x)*g(x) = (3x + 2)(x^2 + 1) = 3x^3 + 2x^2 + 3x + 2

(f/g)(x) = f(x)/g(x) = (3x + 2)/(x^2 + 1)

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