Exam2_2 6 - Explanation ±or a function of the form F x = x-a x-b x-m we see that F a = F b in addition since the denominator is zero only at x = m

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Taylor, Douglas – Exam 2 – Due: Oct 31 2007, 1:00 am – Inst: JEGilbert 6 1. ( -∞ , - 1) , ( - 1 , 0] , [2 , ) 2. [ - 2 , 0] 3. ( -∞ , 0] , [2 , ) 4. ( - 1 , 1) 5. [0 , 2] correct 6. ( -∞ , - 2] , [0 , ) 7. ( -∞ , - 2] , [0 , 1) , (1 , ) Explanation: By the Quotient Rule, f 0 ( x ) = 2 x ( x + 1) 3 - 3 x 2 ( x + 1) 2 ( x + 1) 6 = 2 x ( x + 1) - 3 x 2 ( x + 1) 4 = x (2 - x ) ( x + 1) 4 . Now f will be increasing on an interval [ a, b ] (resp. [ a, b )) when f 0 ( x ) > 0 on the interval ( a, b ) (resp. [ a, b ) if f ( b ) is not deFned). In this case the inequality f 0 ( x ) > 0 will hold when x (2 - x ) > 0 , x 6 = - 1 . Consequently, f will be increasing on [0 , 2] . keywords: increasing, decreasing, rational function, Frst derivative test 013 (part 1 of 1) 10 points Determine if Rolle’s Theorem can be ap- plied to f ( x ) = x 2 + 3 x - 18 x + 9 on the interval [ - 6 , 3], and if it can, Fnd all numbers c satisfying the conclusion of that theorem. 1. c = - 3 , - 3 2 2. Rolle’s Theorem not applicable 3. c = - 1 4. c = - 3 2 5. c = - 3 , - 15 6. c = - 3 correct
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Unformatted text preview: Explanation: ±or a function of the form F ( x ) = ( x-a )( x-b ) x-m we see that F ( a ) = F ( b ); in addition, since the denominator is zero only at x = m , F is continuous and di²erentiable on (-∞ , m ) [ ( m, ∞ ) . Thus Rolle’s Theorem can be applied to F on the interval [ a, b ] so long as m does not belong to ( a, b ). When f ( x ) = ( x + 6)( x-3) x + 9 , therefore, Rolle’s Theorem applies to f on the interval [-6 , 3]. Now, by the Quotient Rule, f ( x ) = (2 x + 3)( x + 9)-( x 2 + 3 x-18) ( x + 9) 2 = x 2 + 18 x + 45 ( x + 9) 2 . But x 2 + 18 x + 45 = ( x + 3)( x + 15) , so f (-3) = 0 = f (-15) ....
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This note was uploaded on 02/14/2012 for the course MATH 408 K taught by Professor Clark,c.w./hoy,r.r during the Spring '08 term at University of Texas at Austin.

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