2023W1_MATH_100C_ALL_2023W1.RMS560AP9L04.WW4

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Feb 20, 2024

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Sayra Arij 2023W1 MATH 100C ALL 2023W1 Assignment WW4 due 10/06/2023 at 11:59pm PDT Problem 1. (1 point) () If f ( x ) = 2 x 8 - 3 x 5 - 5 x 3 + 3 x , find f 0 ( x ) and f 0 ( 1 ) . f 0 ( x ) = f 0 ( 1 ) = Answer(s) submitted: 16 x 7 - 15 x 4 - 15 x 2 + 3 • - 11 submitted: (correct) recorded: (correct) Correct Answers: 16 x 7 - 15 x 4 - 15 x 2 + 3 • - 11 Problem 2. (1 point) () Let f ( x ) = 2 x 2 - 2 x + 7. Compute f 0 ( 2 ) . Answer: Use this to find the equation of the tangent line to the parabola y = 2 x 2 - 2 x + 7 at the point ( 2 , 11 ) . The equation of this tangent line can be written in the form y = mx + b . Determine m and b . m = b = Answer(s) submitted: 6 6 • - 1 submitted: (correct) recorded: (correct) Correct Answers: 6 6 • - 1 Problem 3. (1 point) () Differentiate: Y ( u ) = ( u - 2 + u - 3 )( u 5 - 2 u 2 ) Y 0 ( u ) = Answer(s) submitted: u 5 - 2 u 2 - 2 u 3 - 3 u 4 + 1 u 2 + 1 u 3 5 u 4 - 4 u submitted: (correct) recorded: (correct) Correct Answers: - 2 u - 3 - 3 u - 4 u 5 + - 2 u 2 + u - 2 + u - 3 5 u 4 + 2 · ( - 2 ) u Problem 4. (1 point) () Differentiate: F ( y ) = 1 y 2 - - 9 y 4 ( y - 3 y 3 ) F 0 ( y ) = Answer(s) submitted: y - 3 y 3 - 2 y - 3 - 36 y - 5 + 1 y 2 + 9 y 4 1 - 9 y 2 submitted: (correct) recorded: (correct) Correct Answers: • - 3 + - 9 · ( - 3 ) - 1 y 2 + 3 · ( - 9 ) y 4 Problem 5. (1 point) () Let f ( x ) = 2 x 3 ( x 2 - 7 ) . Evaluate f 0 ( x ) at the following points: (A) f 0 ( 3 ) = (B) f 0 ( - 7 ) = Answer(s) submitted: 432 21952 submitted: (correct) recorded: (correct) Correct Answers: 3 · 2 · 3 2 3 2 - 7 + 2 · 3 3 · 2 · 3 3 · 2 · ( - 7 ) 2 ( - 7 ) 2 - 7 + 2 · ( - 7 ) 3 · 2 · ( - 7 ) 1
Problem 6. (1 point) () Find the equation of the tangent line to the curve y = x x at the point ( 9 , 27 ) . y = Answer(s) submitted: 9 2 x - 27 2 submitted: (correct) recorded: (correct) Correct Answers: 3 2 · 9 1 2 ( x - 9 )+ 27 Problem 7. (1 point) () Find the parabola with equation y = ax 2 + bx whose tangent line at ( 1 , 5 ) has equation y = 7 x - 2. a = b = Answer(s) submitted: 2 3 submitted: (correct) recorded: (correct) Correct Answers: 2 7 - 4 Problem 8. (1 point) () For what value(s) of x is the tangent line of the graph of f ( x ) = 10 x 3 + 75 x 2 + 119 x + 30 parallel to the line y = 1 . 9 - x ? x = (If there is more than one value enter your answer as a comma separated list, eg. ”2,3,4” without quotes.) Answer(s) submitted: • - 1 , - 4 submitted: (correct) recorded: (correct) Correct Answers: • - 4 , - 1 Problem 9. (1 point) () Differentiate: g ( x ) = 8 x - 1 2 x + 5 g 0 ( x ) = Answer(s) submitted: 8 ( 2 x + 5 ) - 2 ( 8 x - 1 ) ( 2 x + 5 ) 2 submitted: (correct) recorded: (correct) Correct Answers: 8 · 5 + 2 ( 2 x + 5 ) 2 Problem 10. (1 point) () Find the derivative of F ( x ) = x - 10 x x x . Note that you can do this two ways; either by simplifying first, or by using the Quotient Rule. Both approaches should obtain the same answer. F 0 ( x ) = Answer(s) submitted: 1 2 x - 1 2 - 10 submitted: (correct) recorded: (correct) Correct Answers: 1 2 x - 1 2 - 10 Problem 11. (1 point) () Differentiate the following function: R ( x ) = 2 x 5 R 0 ( x ) = Answer(s) submitted: • - 5 2 x - 6 submitted: (correct) recorded: (correct) Correct Answers: • - 2 · 5 x - 5 - 1 2
Problem 12. (1 point) () Let f ( x ) = x x + 5 x . Find f 0 ( x ) . f 0 ( x ) = Answer(s) submitted: 10 x ( x 2 + 5 ) 2 submitted: (correct) recorded: (correct) Correct Answers: x + 5 x - x 1 - 5 x 2 x + 5 x 2 Problem 13. (1 point) () Compute the derivative of the given function. f ( x ) = ( 5 x 2 - 7 x + 9 ) 10 x - 1 5 x 2 - 7 x + 9 f 0 ( x ) = . Answer(s) submitted: 10 submitted: (correct) recorded: (correct) Correct Answers: 10 Problem 14. (1 point) () Let f ( x ) = x - 6 x + 6 . Find f 0 ( x ) . f 0 ( x ) = Find f 0 ( 5 ) . f 0 ( 5 ) = Answer(s) submitted: 6 ( x + 6 ) 2 x 6 5 + 6 2 5 submitted: (correct) recorded: (correct) Correct Answers: 1 2 x ( x + 6 ) - ( x - 6 ) 1 2 x ( x + 6 ) 2 0 . 0395573 Problem 15. (1 point) () Compute the derivatives of the following functions: (a) The Michaelis-Menten kinetics function v = Kx k + x . dv dx = (b) The Hill function y = Ax n a n + x n . dy dx = Answer(s) submitted: Kk ( k + x ) 2 na n Ax n - 1 ( a n + x n ) 2 submitted: (correct) recorded: (correct) Correct Answers: Kk ( k + x ) 2 Anx n - 1 a n ( a n + x n ) 2 Problem 16. (1 point) () Suppose that h ( x ) = f ( x ) g ( x ) and f ( - 4 ) = - 4 , g ( - 4 ) = 4 , f 0 ( - 4 ) = 4 , and g 0 ( - 4 ) = 8 . Find h 0 ( - 4 ) . h 0 ( - 4 ) = Answer(s) submitted: 3 submitted: (correct) recorded: (correct) Correct Answers: 4 · 4 -- 4 · 8 4 2 3
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Problem 17. (1 point) () If f ( 5 ) = - 6, g ( 5 ) = 8, f 0 ( 5 ) = 4, and g 0 ( 5 ) = - 4, find the values of the following: (a) ( fg ) 0 ( 5 ) = (b) ( f / g ) 0 ( 5 ) = (c) ( g / f ) 0 ( 5 ) = Answer(s) submitted: 56 1 8 - 8 36 submitted: (correct) recorded: (correct) Correct Answers: 56 8 64 - 8 36 Problem 18. (1 point) () The following table gives the values for functions f and g and their derivatives for integer values of x between 1 and 5. x 1 2 3 4 5 f ( x ) 2 5 -3 -2 -3 f 0 ( x ) -4 -5 1 2 5 g ( x ) 3 -2 -3 1 -1 g 0 ( x ) -1 -2 -2 -1 -5 Let p ( x ) = f ( x ) g ( x ) , q ( x ) = f ( x ) g ( x ) r ( x ) = x f ( x )+ g ( x ) x Find a) p 0 ( 2 ) = b) q 0 ( 3 ) = c) r 0 ( 4 ) = Answer(s) submitted: 0 • - 1 91 16 submitted: (correct) recorded: (correct) Correct Answers: 0 • - 1 5 . 6875 Problem 19. (1 point) Note: You can click on the graph to obtain a larger version in a new browser window. () The graphs of the function f (given in blue, thinner) and g (given in red, thicker) are plotted above. Suppose that u ( x ) = f ( x ) g ( x ) and v ( x ) = f ( x ) / g ( x ) . Find each of the following: u 0 ( 1 ) = v 0 ( 1 ) = Answer(s) submitted: 3 4 21 16 submitted: (correct) recorded: (correct) Correct Answers: 0 . 75 1 . 3125 4
Problem 20. (1 point) () By applying the Product Rule twice, one can prove that if f , g , and h are differentiable, then ( fgh ) 0 = f 0 gh + fg 0 h + fgh 0 . Now, in the above result, letting f = g = h yields d dx [ f ( x )] 3 = 3 [ f ( x )] 2 f 0 ( x ) . Use this last formula to differentiate y = e 3 x . y 0 = Answer(s) submitted: 3 e 3 x submitted: (correct) recorded: (correct) Correct Answers: 3 e 3 x Problem 21. (1 point) () Suppose that f ( x ) = 16 e x - ex e . Find f 0 ( 3 ) . f 0 ( 3 ) = Answer(s) submitted: 16 e 3 - e 2 · 3 e - 1 submitted: (correct) recorded: (correct) Correct Answers: 272 . 569 Problem 22. (1 point) () Let f ( x ) = - 5 e x + 2 + e 5 . f 0 ( 0 ) = Answer(s) submitted: • - 5 e 2 submitted: (correct) recorded: (correct) Correct Answers: • - 36 . 9452804946533 Problem 23. (1 point) () If f ( x ) = x 3 + 3 e x , find f 0 ( 5 ) . f 0 ( 5 ) = Use this to find the equation of the tangent line to the curve y = x 3 + 3 e x at the point ( a , f ( a )) when a = 5. The equation of this tangent line can be written in the form y = mx + b . Find m = and b . m = b = Answer(s) submitted: 75 + 3 e 5 75 + 3 e 5 • - 250 + 12 e 5 submitted: (correct) recorded: (correct) Correct Answers: 520 . 23947730773 520 . 23947730773 • - 2030 . 95790923092 Problem 24. (1 point) () Find the equation of the tangent line to the curve y = 11 xe x at the point ( 0 , 0 ) . The equation of this tangent line can be written in the form y = mx + b . Find m and b . m = b = Answer(s) submitted: 11 0 submitted: (correct) recorded: (correct) Correct Answers: 11 0 5
Problem 25. (1 point) () Consider f ( x ) = 5 - e x . A. Find the slope of the graph of f ( x ) at the point where the graph crosses the x -axis. slope = B. Find the equation of the tangent line to the curve at this point. y = C. Find the equation of the line perpendicular to the tangent line at this point. (This is the normal line.) y = Answer(s) submitted: • - e ln ( 5 ) • - e ln ( 5 ) ( x - ln ( 5 )) 1 e ln ( 5 ) ( x - ln ( 5 )) submitted: (correct) recorded: (correct) Correct Answers: • - 1 · 5 • - 1 · 5 ( x - ln ( 5 )) 1 5 ( x - ln ( 5 )) Generated by ©WeBWorK, http://webwork.maa.org, Mathematical Association of America 6
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