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Stitz-Zeager_College_Algebra_e-book

# 4 we dened a function as a special type of relation

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Unformatted text preview: = x2 − x − 2 with the quantity (x + h) to get f (x + h) = (x + h)2 − (x + h) − 2 = x2 + 2xh + h2 − x − h − 2. So the diﬀerence quotient is f (x + h) − f (x) h 2 = x2 + 2xh + h2 − x − h − 2 − x2 − x − 2 h You may need to brush up on your Intermediate Algebra skills, as well. 1.6 Function Arithmetic 59 x2 + 2xh + h2 − x − h − 2 − x2 + x + 2 h 2−h 2xh + h h h (2x + h − 1) h h  (2x + h − 1) h  = = = = = factor cancel 2x + h − 1. 2. To ﬁnd g (x + h), we replace every occurrence of x in the formula g (x) = quantity (x + h) g (x + h) = 3 2(x + h) + 1 = 3 with the 2x + 1 3 , 2x + 2 h + 1 which yields g (x + h) − g (x) h = = 3 3 − 2x + 2 h + 1 2x + 1 h 3 3 − 2x + 2h + 1 2x + 1 · (2x + 2h + 1)(2x + 1) h (2x + 2h + 1)(2x + 1) = 3(2x + 1) − 3(2x + 2h + 1) h(2x + 2h + 1)(2x + 1) = 6x + 3 − 6x − 6h − 3 h(2x + 2h + 1)(2x + 1) = −6h h(2x + 2h + 1)(2x + 1) = −6 h h (2x + 2h + 1)(2x + 1) = −6 . (2x + 2h + 1)(2x + 1) For reasons which will become clear in Cal...
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