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Answer :
Answer:
f(x) = (2x+3)/(x+1) and g(x) = x/(x-1)
f+g = f(x) + g(x) = (2x+3)/(x+1) + x/(x-1)
To add the two functions, we need a common denominator:
lcm(x+1,x-1) = (x+1)(x-1)
So we can write:
f+g = [(2x+3)(x-1) + (x)(x+1)]/[(x+1)(x-1)]
f+g = [(2x^2 + x - 3) + (x^2+x)]/[(x+1)(x-1)]
f+g = (3x^2 + 2x - 3)/[(x+1)(x-1)]
The domain of f+g is all x such that (x+1)(x-1) ≠ 0 or x ≠ -1 and x ≠ 1. The domain can be written in interval notation as (-∞,-1) ∪ (-1, 1) ∪ (1, ∞).
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f−g = f(x) - g(x) = (2x+3)/(x+1) - x/(x-1)
To subtract the two functions, we need a common denominator:
lcm(x+1,x-1) = (x+1)(x-1)
So we can write:
f-g = [(2x+3)(x-1) - (x)(x+1)]/[(x+1)(x-1)]
f-g = [(2x^2 - x - 3) - (x^2+x)]/[(x+1)(x-1)]
f-g = (x^2 - 3x - 3)/[(x+1)(x-1)]
The domain of f-g is all x such that (x+1)(x-1) ≠ 0 or x ≠ -1 and x ≠ 1. The domain can be written in interval notation as (-∞,-1) ∪ (-1, 1) ∪ (1, ∞).
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f⋅g = f(x) * g(x) = (2x+3)/(x+1) * x/(x-1)
f⋅g = (2x^2 + 3
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