# hw9so - Create assignment 57321 Homework 9 Apr 05 at 9:08...

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Create assignment, 57321, Homework 9, Apr 05 at 9:08 pm 1 This print-out should have 19 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. The due time is Central time. CalC7a01b 52:03, calculus3, multiple choice, < 1 min, fxed. 001 Use f 0 ( x ) to determine whether f ( x ) = x 4 4 - 2 x 2 has an inverse on ( -∞ , ). 1. f does not have inverse correct 2. f has inverse Explanation: The Function will have an inverse iF f is in- creasing everywhere on ( -∞ , ) or decreas- ing everywhere, so we need to look at the sign oF f 0 ( x ). Now f 0 ( x ) = x 3 - 4 x = x ( x 2 - 4) , and so has sign chart -∞ - 2 0 2 - + - + Thus f is decreasing on ( -∞ , - 2] and on [0 , 2], but increasing on [ - 2 , 0] and on [2 , ). Consequently, f does not have an inverse . keywords: inverse Functions CalC7a27s 52:03, calculus3, multiple choice, > 1 min, wording-variable. 002 ±ind the inverse Function, f - 1 , oF f when f is defned by f ( x ) = 5 x - 6 , x 6 5 . 1. f - 1 ( x ) = 1 5 ( x 2 + 6) , x 0 correct 2. f - 1 ( x ) = 1 5 ( x 2 + 6) , x 5 6 3. f - 1 ( x ) = 1 5 p x 2 - 6 , x 0 4. f - 1 ( x ) = 1 6 ( x 2 - 5) , x 6 5 5. f - 1 ( x ) = 1 6 p x 2 + 5 , x 5 6 6. f - 1 ( x ) = 1 6 p x 2 - 5 , x 0 Explanation: Since f has domain [ 6 5 , ) and is increasing on its domain, the inverse oF f exists and has range [ 6 5 , ); Furthermore, since f has range [0 , ), the inverse oF f has domain [0 , ). To determine f - 1 we solve For x in y = 5 x - 6 and then interchange x, y . Solving frst For x , we see that 5 x = y 2 + 6 . Consequently, f - 1 is defned on [0 , ) by f - 1 ( x ) = 1 5 ( x 2 + 6) . keywords: CalC7a41s 52:03, calculus3, multiple choice, < 1 min, wording-variable. 003 On ( - 1 , 1) the Function f ( x ) = 5 + x 2 + tan πx 2 ·

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Create assignment, 57321, Homework 9, Apr 05 at 9:08 pm 2 has an inverse g . Find the value of g 0 (5). (Hint: fnd the value oF f (0)). 1. g 0 (5) = 2 π correct 2. g 0 (5) = π 2 3. g 0 (5) = 1 4. g 0 (5) = 5 π 5. g 0 (5) = π 5 Explanation: By de±nition of inverse function: f ( g ( x )) = x, g ( f ( x )) = x. Applying the Chain Rule to the ±rst of these equations, we see that f 0 ( g ( x )) g 0 ( x ) = 1 , in which case g 0 ( x ) = 1 f 0 ( g ( x )) . Thus to ±nd the value of g 0 (5) we need to know the value of g (5) as well as f 0 ( x ). Now f 0 ( x ) = 2 x + π 2 sec 2 πx 2 · . The hint becomes useful in determining the value of g (5). For f (0) = 5, and so by the second of the equations above, g (5) = g ( f (0)) = 0 ,. Consequently,
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## This note was uploaded on 03/20/2008 for the course M 408c taught by Professor Mcadam during the Spring '06 term at University of Texas.

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hw9so - Create assignment 57321 Homework 9 Apr 05 at 9:08...

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