# 231003Worksheet 3.1

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Prepared by Ryuichi Man 1 MATH115 Calculus I (Section 129) Fall 2023 Section 3.1 Worksheet Problem 1 : A portion of the graph of the function ( ) m x , which is defined on ( ) , −  , is shown on the right. Note that ( ) m x is linear on the intervals ( ) 4, 1 , ( ) 1,2 and ( ) 2,6 . Find the exact values, or write DNE if the value does not exist. a. Let ( ) ( ) 3 2 P x m x x = . Find ( ) ' 3 P . b. Let ( ) ( ) 3 2 Q b b m b = + . Find ( ) " 5 Q . c. Let ( ) ( ) 6 R t t t m t = + . Find ( ) ' 2 R . d. Let ( ) ( ) 2 5 3 u u S u m u u = . Find ( ) ' 1 S . Solution : a. ( ) ( ) 2 3 ' 2 ' P x m x x = − ( ) ( ) ( ) 2 3 1 13 ' 3 2 ' 3 2 2 3 3 3 P m = − = − = − b. ( ) ( ) 2 ' 3 2 ' Q b b m b = + ( ) ( ) " 6 2 " Q b b m b = + ( ) ( ) ( ) ( ) " 5 6 5 2 " 5 30 2 0 30 Q m = + = + = [Since ( ) m x is linear on ( ) 2,6 , its second derivative is always zero on ( ) 2,6 .] c. ( ) ( ) 1 1 ' 6 ' 2 R t t m t t = + Since ( ) m x is not differentiable at 2 x = , we conclude that ( ) ' 2 R does not exist ( DNE ). d. ( ) ( ) 3 1 2 5 3 S u u u m u = ( ) ( ) 5 2 2 3 ' 5 3 ' 2 S u u u m u = + ( ) ( ) ( ) ( ) 5 2 2 3 3 2 3 ' 1 5 1 3 ' 1 5 3 2 2 3 2 S u m = + = − + =
Prepared by Ryuichi Man 2 Problem 2 : Let ( ) p x be a differentiable function defined on ( ) , −  . Define ( ) ( ) 2 3 q x x p x = . If the equation of the tangent line to ( ) y p x = at the point 1 x = is 2 5 y x = , find the equation of the tangent line to ( ) y q x = at the point 1 x = . Solution : Note that ( ) ( ) 1 2 1 5 3 p = = − and ( ) ' 1 2 p = . Thus, we have ( ) ( ) 2 1 1 3 1 10 q p = = and ( ) ( ) ( ) ' 1 2 1 3 ' 1 4 q p = = − . Thus, the required equation is given by ( ) 4 1 10 y x = − + , or 4 14 y x = − + . Problem 3 : [Modified from 2018 Fall Exam 2 Problem 2] Let ( ) 2 3 3 0 1 1 B x x k x x Bx Ax x + = + where A and B be constants. a. Find the values of A and B such that ( ) k x is differentiable on ( ) 0, . If there are no such values of A and B , write NONE . b. Is ( ) ' k x differentiable at 1 x = ? Explain your answer. Solution : a. Since ( ) k x is differentiable at 1 x = , ( ) k x is continuous at 1 x = . Thus, we have ( ) ( ) ( ) 2 3 3 1 1 1 1 B B A + = + 3 A = . Note that ( ) 2 2 3 0 1 ' 2 9 1 B x k x x Bx x x = + . Since ( ) k x is differentiable at 1 x = , we have ( ) ( ) ( ) 2 2 3 2 1 9 1 1 B B = + 2 B = − . b. Note that ( ) 3 4 0 1 " 4 18 1 x k x x x x = + . Since ( ) ( ) 3 4 4 14 4 18 1 1 = − = − + , we conclude that ( ) ' k x is not differentiable at 1 x = .