Test2_solutions

# Test2_solutions - Math 1432 Exam 2 Review KEY 1 f x = 3x 2...

This preview shows pages 1–3. Sign up to view the full content.

Math 1432 Exam 2 Review -­‐ KEY 1. a. 2 1 1/3 '( ) 3 0 for all x => f(x) is always increasing => one-to-one ( ) ( 1) f x x f x x = = b. 1 '( ) 3 0 for all x => f(x) is always increasing => one-to-one 10 ( ) 3 f x x f x = > = c. 2 '( ) => not one-to-one 9 x f x x = 2. 1 7 ( )'(1) 2 f = . 3. 2 7 4. 1 12 5. Find the derivative: a. ( ) 4 ' 2 4 x x e y e x + = + b. ( ) 6 6 ' cos ln(5 ) 5 y x x = c. 2 2 2 ' 2 2 2 x x y xe x e = + + d. 2 2 ' 2 cosh(3 ) 3 sinh(3 ) x x y xe x e x = + e. tan '( ) 2 x f x x = f. 4 6 '( ) 2 x x xe xe f x x x = g. [ ] ( 7) ' (cos ) ln(cos ) ( 7)tan x y x x x x + = h. 2 6 6 18 '( ) (3 1) 2ln(3 1) 3 1 x x f x x x x + + = + i. 6 2 2 2 '( ) 6 1 (5 2 ) x f x e x x = + +

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
j. 2 '( ) ln7 f x x = k. ( )( ) 2 ' 2 ln6 6 x y = l. 2 6 6 '( ) 1 4 x f x x = + 6. Integrate: a. 4 1 ln 4 e e dx x = b. 2 9 9 csc 1 1 ln | 2 5cot | 2 5cot 5 9 x x x e dx x e C x = + + c. 2 sinh 1 (2 cosh ) 2 cosh x dx C x x = + + + d. 2 x x e dx e C x = + e. ( ) 2 4ln |3 | 3 dx x C x x = + f. 2 ln | 1| 1 x dx x x C x + = + + + + g. 2 2 2 3 3 3 3 3 ln( 1) 1 2 x x dx x x C x + + = + + + + h. 3 2 2 cos sin sin tan cos x x dx x x x C x = + + i. 1 tan(3 ) ln | sec(3 ) | 3 x dx x C = + j. 2 2 arctan(3 ) 1 (arctan(3 )) 1 9 6 x dx x C x = + + k. 3 2 2 0 1 3 1 dx x π = l. 4 3 5 7 1 1 cos sin cos cos 5 7 x xdx x x C = + + m. 5 2 3 5 7 1 2 1 cos sin sin sin sin 3 5 7 x xdx x x x C = + + n. 3 2 1 cot cot ln | sin | 2 xdx x x C = + o. 2 2 1 1 ln(2 ) ln(2 )
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern