m2121t4rev-sol-fs2016

# m2121t4rev-sol-fs2016 - MATH/2121 CALCULUS FOR THE LIFE...

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MATH/2121, CALCULUS FOR THE LIFE SCIENCES I TEST 4 REVIEW PROBLEMS SOLUTIONS (1) Here f ( x ) = x 3 - 6 x + 4. (a) We have f 0 ( x ) = 3 x 2 - 6 = 3( x 2 - 2). So f 0 ( x ) exists for all x and f 0 ( x ) = 0 x 2 = 2 x = ± 2 . Critical numbers : x = - 2 , 2. (b) The critical numbers partition the real line into three open intervals: ( -∞ , - 2) , ( - 2 , 2) , ( 2 , ) . We pick a test point x in each interval and check the sign of f 0 ( x ): For interval ( -∞ , - 2) we take test point x = - 2 and compute f 0 ( - 2) = 6 > 0. For interval ( - 2 , 2) we take test point x = 0 and compute f 0 (0) = - 6 < 0. For interval ( 2 , ) we take test point x = 2 and compute f 0 (2) = 6 > 0. Intervals of increase/decrease : f is increasing on ( -∞ , - 2) and ( 2 , ). It is decreasing on ( - 2 , 2). (c) f 00 ( x ) = 6 x has f 00 ( - 2) = - 6 2 < 0 , - - f 00 ( 2) = 6 2 > 0 + + The second derivative test ensures that we have a relative maximum at x = - 2 and a relative minimum at x = 2. We compute f ( - 2) = ( - 2) 3 - 6( - 2) + 4 = - 2 2 + 6 2 + 4 = 4 + 4 2 , f ( 2) = ( 2) 3 - 6( 2) + 4 = 2 2 - 6 2 + 4 = 4 - 4 2 . Relative maxima : ( - 2 , 4 + 4 2), Relative minima : ( 2 , 4 - 4 2) Note : Graphs for the functions in problems #1-6 can be found at the end of these solutions. (2) Here f ( x ) = x 4 - 6 x 2 + 9. 1

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(a) We have f 0 ( x ) = 4 x 3 - 12 x = 4 x ( x 2 - 3). So f 0 ( x ) exists for all x and f 0 ( x ) = 0 x = 0 or x 2 = 3 x = 0 or x = ± 3 . Critical numbers : x = - 3 , 0 , 3. (b) The critical numbers partition the real line into four open intervals: ( -∞ , - 3) , ( - 3 , 0) , (0 , 3) , ( 3 , ) . We pick a test point x in each interval and check the sign of f 0 ( x ): For interval ( -∞ , - 3) we take test point x = - 3 and compute f 0 ( - 3) = - 72 < 0. For interval ( - 3 , 0) we take test point x = - 1 and compute f 0 ( - 1) = 8 > 0. For interval (0 , 3) we take test point x = 1 and compute f 0 (1) = - 8 < 0. For interval ( 3 , ) we take test point x = 3 and compute f 0 (3) = 72 > 0. Intervals of increase/decrease : f is increasing on ( - 3 , 0) and ( 3 , ). It is decreasing on ( -∞ , - 3) and (0 , 3). (c) f 00 ( x ) = 12 x 2 - 12 has f 00 ( - 3) = 12 × 3 - 12 = 24 > 0 , + + f 00 (0) = - 12 < 0 , - - f 00 ( 3) = 12 × 3 - 12 = 24 > 0 + + The second derivative test ensures that we have a relative mimima at x = ± 3 and a relative maximum at x = 0. We compute f ( - 3) = ( - 3) 4 - 6( - 3) 2 + 9 = 9 - 18 + 9 = 0 , f (0) = 9 , f ( 3) = ( 3) 4 - 6( 3) 2 + 9 = 9 - 18 + 9 = 0 . Relative maxima : (0 , 9), Relative minima : ( - 3 , 0) and ( 3 , 0). (3) Here f ( x ) = - 2 / ( x 2 + 9) = - 2( x 2 + 9) - 1 . 2
(a) We compute f 0 ( x ) = ( - 2)( - 1)( x 2 + 9) - 2 (2 x ) = 4 x ( x 2 + 9) - 2 . So f 0 ( x ) exists for all x and f 0 ( x ) = 0 x = 0 . Critical numbers : x = 0. (b) The critical number partitions the real line into two open intervals: ( -∞ , 0) , (0 , ) . We pick a test point x in each interval and check the sign of f 0 ( x ): For interval ( -∞ , 0) we take test point x = - 1 and compute f 0 ( - 1) = - 4 / 100 < 0. For interval (0 , ) we take test point x = 1 and compute f 0 (1) = 4 / 100 > 0. Intervals of increase/decrease : f is decreasing on ( -∞ , 0). It is increasing on (0 , ).

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