Precalc0110to0111-page7

Precalc0110to0111-page7 - (Section 1.10: Difference...

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(Section 1.10: Difference Quotients) 1.10.7 Example 3 (Form 2 of a Difference Quotient; Revisiting Example 2 on Profit) Again, Px () = ± x 2 + 200 x ± 5000 . Evaluate ± P 60 x ± 60 . (Form 2) § Solution ± P 60 x ± 60 = ± x 2 + 200 x ± 5000 ² ³ ´ µ ±± 60 2 + 200 60 ± 5000 ² ³ ´ µ · x ± 60 = ± x 2 + 200 x ± 5000 ² ³ ´ µ 3600 + 12,000 ± 5000 ² ³ ´ µ x ± 60 = ± x 2 + 200 x + 3600 ± 12,000 x ± 60 = ± x 2 + 200 x ± 8400 x ± 60 = ± x 2 ± 200 x + 8400 x ± 60 • The Factor Theorem in Chapter 2 will imply that,
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