231003Worksheet 3.1

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 = .