Stitz-Zeager_College_Algebra_e-book

I f g x ii g hx f x g iv f hx iii v

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Unformatted text preview: − f )(x) is defined to be g (x) − f (x). To that end, we get (g − f )(x) = g (x) − f (x) = 3− 1 x − 6x2 − 2x 1 − 6x2 + 2x x 3x 1 6x3 2x2 −− + x x x x 3 + 2 x2 + 3 x − 1 −6x x = 3− = = get common denominators Looking at the expression for (g − f ) before we simplified 3− 1 x − 6x2 − 2x we see, as before, x = 0 is the only restriction. The domain is (−∞, 0) ∪ (0, ∞). 1.6 Function Arithmetic 57 3. (f g )(x) is defined to be f (x)g (x). Substituting yields (f g )(x) = f (x)g (x) 1 x = 6x2 − 2x 3− = 6x2 − 2x 3x − 1 x = 2x(3x − 1) 1 3x − 1 x factor = 2&(3x − 1) x 1 3x − 1 x & cancel = 2(3x − 1)2 = 2 9x2 − 6x + 1 = 18x2 − 12x + 2 To determine the domain, we check the step just after we substituted 6x2 − 2x 3− 1 x which gives us, as before, the domain: (−∞, 0) ∪ (0, ∞). 4. g f (x) is defined to be g f (x) g (x) . Thus we have f (x) = g (x) f (x) 1 x 6x2 − 2x 1 3− x ·x 2 − 2x x 6x 1 3− x x (6x2 − 2x) x 3...
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This note was uploaded on 05/03/2013 for the course MATH Algebra taught by Professor Wong during the Fall '13 term at Chicago Academy High School.

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