Stitz-Zeager_College_Algebra_e-book

# Hx 5 4 3x 3x2 x solution 1 in the expression for f

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Unformatted text preview: 1)2 + 3(−1) + 4 = −(1) + (−3) + 4 =0 Similarly, f (0) = −(0)2 + 3(0) + 4 = 4, and f (2) = −(2)2 + 3(2) + 4 = −4 + 6 + 4 = 6. 2. To ﬁnd f (2x), we replace every occurrence of x with the quantity 2x f (2x) = −(2x)2 + 3(2x) + 4 = −(4x2 ) + (6x) + 4 = −4x2 + 6x + 4 The expression 2f (x) means we multiply the expression f (x) by 2 2f (x) = 2 −x2 + 3x + 4 = −2x2 + 6x + 8 46 Relations and Functions Note the diﬀerence between the answers for f (2x) and 2f (x). For f (2x), we are multiplying the input by 2; for 2f (x), we are multiplying the output by 2. As we see, we get entirely diﬀerent results. Also note the practice of using parentheses when substituting one algebraic expression into another; we highly recommend this practice as it will reduce careless errors. 3. To ﬁnd f (x + 2), we replace every occurrence of x with the quantity x + 2 f (x + 2) = = = = −(x + 2)2 + 3(x + 2) + 4 − x2 + 4x + 4 + (3x + 6) + 4 −x2 − 4x − 4 + 3x + 6 + 4 −x2 − x + 6 To ﬁnd f (x) + 2, we add 2 to the expression for f (x) f (x) + 2 = −x2 + 3x + 4 +...
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