Math 115 Winter 2017, Written Assignment 2 Due by Wednesday February 8, 5:00 pm 1. Given that f (x) = 2x2 8x + 9, 0 x 3. (a) Find the average value of f . (b) Find all c with fave = f (c). 2. Find the values of the following. (a) sin(2 cos1 ( 15 )) (b) cos(2 tan1 ( 3)) (c) sin2 ( 12 tan1 ( 73 )) 3. Find the values of the following. (a) cot(sin1 ( 12 ) sec1 (2)), (b) cos(tan1 (2) + tan1 (3)) 4. Evaluate the definite integrals. Simplify your answers as much as possible. Z ln 2 Z 3/2 Z 2 e2x ln(x + 2) dx dx (b) dx (c) (a) 4x x+2 e +3 4x2 + 9 0 1 0 Z 1+x (d) dx 3 x2 5. Find the volume of a solid generated when the region bounded by the curves y = ex , x = 0, x = ln 2 and y = 0 is revolved about the line y = 1. Include a diagram showing the region and a typical disk/washer. 6. (a) Show that f (x) = 1 , x > 1 is one-to-one. x1 (b) Find (f 1 )0 (2) using the formula proved in class. 7. Suppose g is the inverse function of a differentiable function f and G(x) = 1 . If g(x) 1 f (3) = 2 and f 0 (3) = , find G0 (2). 9 8. (a) Find y 0 if y = tan1 (x x2 + 1). Simplify your answer. (b) Find y 0 (x) and y 0 ( 12 ) if y(x) = (cos1 x)ln(2x+1) 9. Find an equation of the tangent line to the curve 1 sin1 (x2 y) = xy 2 + tan y at the point (0, 4 ). : 1 ( )=2 8 +9, 0 3 ( ) (3) (0) 3 9 = = 2 30 3 = ( ) = ( ) => 2 = 2 => 2 = 8 +9 8 + 11 = 0 8 64 88 8 24 8 2 6 4 6 = = = 4 4 4 2 2 ( ) 1 5 2 1 5 = 2 1 = 5 1 5 1 = 5 => 1 5 2 = => 1 = 5 1 5 = = 2 5 1 5 => 1 24 5 5 24 = 5 24 5 = 224 . 25 ( ) 2 tan tan 2 tan = 3 = 2 3 2 == 5 3 3 , 2 2 tan = 10 3 + 3 3 => 2 tan tan 2 = =? = 3 3 + 3 = 3 2 tan 3 tan = 1 2 5 3 2 tan 3 = 2 = 2 tan 5 6 tan 5 3 = 2 + = = 3 2 2 tan 3 3 = 2 tan = 3 5 6 tan = 1 2 1 2 ( ) 1 tan 2 1 tan 2 7 = 3 => (2 ) = => 7 3 =? 7 3 = tan(2 ) => = 3 49 + 9 3 58 ( )= => 1 2 1 tan 2 7 3 =1 (2 ) 3 58 3 58 3 ( ) 1 2 1 = 2 + (2) = , 2 ,2 3 = = 6 7 11 , 6 6 5 3 3 , 1 (2) 2 7 7 5 11 11 5 = , , , 6 3 6 3 6 3 6 3 = 3 6 (2) = ? 1 2 5 3 , , , 6 2 2 6 (2) = = 5 6 3 3 ,0 ,0 , 2 2 = 3 ,0 . 2 , 2 , 3 2 , 6 ( ) (tan (2) + tan (3)) = = 2+3 16 tan 2 = 4 4 = 1 2 = (tan (1)) 4 4 . 4 ( ) ln( + 2) +2 = ln( + 2) = 1 +2 => = ( ) = ( ) = =? ln(4) 0 1 = [ln(4)] = 2[ln(2)] 2 2 ( ) ( ) = = = 1 2 +3 = => = 1 1 tan 2 3 = =? +3 = 3 2 3 1 2 2 3 4 tan 3 4 tan 3 4 tan 1 3 6 ( ) = 4 1 = +9 4 3 + 2 1 1 = tan 4 3 2 1 = tan 6 1 3 3 2 0 = 36 1 2 = 4 3 tan 2 3 ( ) = 1+ 1 = 3 = = 3 = = = = = 3 = 3 = 3 + 3 3 3 = => 1 3 = =? 3 = + 1+ 3 + = + 3 +( 3 3 + + 3 + + ) + . 5 = , = 0, = ln(2) =0 = 1 . ( ) ( ) ( + 1) = ( ) ( = + 1) +2 ln(2) ( ) = = 2 +2 + ln(2) 1 1 4 + 2 2 + ln(2) + 2 + 0 2 2 = (6 + ln(2)) =6 ln(2) 5 ln(2) 2 5 7 = 2 2 . 6 ( ) ( )= >1 1 , 1 >1 >1 ( )= ( ) 1 = 1 => 1 => 1 1= 1 => ( ) = , ( ) ( )= 1 => 1 1 = 1 => 1 = +11 => + 1 => ( )= 1 => ( +1 )( )= ( ) (2) = 1 1 = . (2 + 1) 9 1 1 ( + 1) 7 ( )= 1 , (3) = 2 , ( ) => (2) = 3 ( (3) ( )= )( ) = ( ) => =3, 1 9 (3) = ( ) ( ) = => ( )=1 (3) = 1 => (2) 1 = 1 => 9 1 [ ( )] (2) = 9 ( )= 1 => ( ) ( )= (2) = 1 [ (2)] 1 (2) = 9 = 1 9 ( ) 8 ( ) = tan => = +1 1 = 1+ + 1 + 1 2 +1 +1 2+2 = 1 = 2+2 2 ) ( ) 2 +1 +1 1 ( ) =( 1 = (1) , = ... ... . (1) +1 +1 1 +1 = +1 => ln( ) = ln(2 + 1) ln{( => 1 1 = ln(2 + 1) =( => ) 1 + ln{( 1 ( )} ln(2 + 1) ) + 1 )} 2 2 +1 )} 2 ln{( 2 +1 = , = 1 = 2 ( ) = 3 1 2 ln ln 2 2 + 1 + 1 1 1 2 2 1 2 2 ln ln(2) 3 + 2 3 4 3 = 23 ln(2) ( ) 3 9 ( 1 => 2 + = = (1 )= + (1 ) 0, 0+0 = 2 , 4 (1 0 1) => ) = 02 16 32 0, 4 => = = 32 32 ( 0) + 4 . 4 1 2 2 1 +1 2 + ln 3 . : 1 ( )=2 8 +9, 0 3 ( ) (3) (0) 3 9 = = 2 30 3 = ( ) = ( ) => 2 = 2 => 2 = 8 +9 8 + 11 = 0 8 64 88 8 24 8 2 6 4 6 = = = 4 4 4 2 2 ( ) 1 5 2 1 5 = 2 1 = 5 1 5 1 = 5 => 1 5 2 = => 1 = 5 1 5 = = 2 5 1 5 => 1 24 5 5 24 = 5 24 5 = 224 . 25 ( ) 2 tan tan 2 tan = 3 = 2 3 2 == 5 3 3 , 2 2 tan = 10 3 + 3 3 => 2 tan tan 2 = =? = 3 3 + 3 = 3 2 tan 3 tan = 1 2 5 3 2 tan 3 = 2 = 2 tan 5 6 tan 5 3 = 2 + = = 3 2 2 tan 3 3 = 2 tan = 3 5 6 tan = 1 2 1 2 ( ) 1 tan 2 1 tan 2 7 = 3 => (2 ) = => 7 3 =? 7 3 = tan(2 ) => = 3 49 + 9 3 58 ( )= => 1 2 1 tan 2 7 3 =1 (2 ) 3 58 3 58 3 ( ) 1 2 1 = 2 + (2) = , 2 ,2 3 = = 6 7 11 , 6 6 5 3 3 , 1 (2) 2 7 7 5 11 11 5 = , , , 6 3 6 3 6 3 6 3 = 3 6 (2) = ? 1 2 5 3 , , , 6 2 2 6 (2) = = 5 6 3 3 ,0 ,0 , 2 2 = 3 ,0 . 2 , 2 , 3 2 , 6 ( ) (tan (2) + tan (3)) = = 2+3 16 tan 2 = 4 4 = 1 2 = (tan (1)) 4 4 . 4 ( ) ln( + 2) +2 = ln( + 2) = 1 +2 => = ( ) = ( ) = =? ln(4) 0 1 = [ln(4)] = 2[ln(2)] 2 2 ( ) ( ) = = = 1 2 +3 = => = 1 1 tan 2 3 = =? +3 = 3 2 3 1 2 2 3 4 tan 3 4 tan 3 4 tan 1 3 6 ( ) = 4 1 = +9 4 3 + 2 1 1 = tan 4 3 2 1 = tan 6 1 3 3 2 0 = 36 1 2 = 4 3 tan 2 3 ( ) = 1+ 1 = 3 = = 3 = = = = = 3 = 3 = 3 + 3 3 3 = => 1 3 = =? 3 = + 1+ 3 + = + 3 +( 3 3 + + 3 + + ) + . 5 = , = 0, = ln(2) =0 = 1 . ( ) ( ) ( + 1) = ( ) ( = + 1) +2 ln(2) ( ) = = 2 +2 + ln(2) 1 1 4 + 2 2 + ln(2) + 2 + 0 2 2 = (6 + ln(2)) =6 ln(2) 5 ln(2) 2 5 7 = 2 2 . 6 ( ) ( )= >1 1 , 1 >1 >1 ( )= ( ) 1 = 1 => 1 => 1 1= 1 => ( ) = , ( ) ( )= 1 => 1 1 = 1 => 1 = +11 => + 1 => ( )= 1 => ( +1 )( )= ( ) (2) = 1 1 = . (2 + 1) 9 1 1 ( + 1) 7 ( )= 1 , (3) = 2 , ( ) => (2) = 3 ( (3) ( )= )( ) = ( ) => =3, 1 9 (3) = ( ) ( ) = => ( )=1 (3) = 1 => (2) 1 = 1 => 9 1 [ ( )] (2) = 9 ( )= 1 => ( ) ( )= (2) = 1 [ (2)] 1 (2) = 9 = 1 9 ( ) 8 ( ) = tan => = +1 1 = 1+ + 1 + 1 2 +1 +1 2+2 = 1 = 2+2 2 ) ( ) 2 +1 +1 1 ( ) =( 1 = (1) , = ... ... . (1) +1 +1 1 +1 = +1 => ln( ) = ln(2 + 1) ln{( => 1 1 = ln(2 + 1) =( => ) 1 + ln{( 1 ( )} ln(2 + 1) ) + 1 )} 2 2 +1 )} 2 ln{( 2 +1 = , = 1 = 2 ( ) = 3 1 2 ln ln 2 2 + 1 + 1 1 1 2 2 1 2 2 ln ln(2) 3 + 2 3 4 3 = 23 ln(2) ( ) 3 9 ( 1 => 2 + = = (1 )= + (1 ) 0, 0+0 = 2 , 4 (1 0 1) => ) = 02 16 32 0, 4 => = = 32 32 ( 0) + 4 . 4 1 2 2 1 +1 2 + ln 3