K Exam 3-solutions

# K Exam 3-solutions - Version 037 – K Exam 3 – Hamrick...

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

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Version 037 – K Exam 3 – Hamrick – (54868) 1 This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Find all functions g such that g ′ ( x ) = 2 x 2 + x + 4 √ x . 1. g ( x ) = 2 √ x ( 2 x 2 + x − 4 ) + C 2. g ( x ) = 2 √ x ( 2 x 2 + x + 4 ) + C 3. g ( x ) = 2 √ x parenleftbigg 2 5 x 2 + 1 3 x − 4 parenrightbigg + C 4. g ( x ) = √ x parenleftbigg 2 5 x 2 + 1 3 x + 4 parenrightbigg + C 5. g ( x ) = √ x ( 2 x 2 + x + 4 ) + C 6. g ( x ) = 2 √ x parenleftbigg 2 5 x 2 + 1 3 x + 4 parenrightbigg + C cor- rect Explanation: After division g ′ ( x ) = 2 x 3 / 2 + x 1 / 2 + 4 x − 1 / 2 , so we can now find an antiderivative of each term separately. But d dx parenleftbigg ax r r parenrightbigg = ax r − 1 for all a and all r negationslash = 0. Thus 4 5 x 5 / 2 + 2 3 x 3 / 2 + 8 x 1 / 2 = 2 √ x parenleftbigg 2 5 x 2 + 1 3 x + 4 parenrightbigg is an antiderivative of g ′ . Consequently, g ( x ) = 2 √ x parenleftbigg 2 5 x 2 + 1 3 x + 4 parenrightbigg + C with C an arbitrary constant. 002 10.0 points Find the value of f (0) when f ′ ( t ) = sin 2 t , f parenleftBig π 2 parenrightBig = − 1 . 1. f (0) = 2 2. f (0) = − 2 correct 3. f (0) = − 1 4. f (0) = 0 5. f (0) = 1 Explanation: Since d dx cos mt = − m sin mt , for all m negationslash = 0, we see that f ( t ) = − 1 2 cos 2 t + C where the arbitrary constant C is determined by the condition f ( π/ 2) = − 1. But cos 2 t vextendsingle vextendsingle vextendsingle t = π/ 2 = cos π = − 1 . Thus f parenleftBig π 2 parenrightBig = 1 2 + C = − 1 , and so f ( t ) = − 1 2 cos 2 t − 3 2 . Consequently, f (0) = − 2 . 003 10.0 points Determine the inverse, f − 1 , of f when f ( x ) = 5 − 3 x 3 . Version 037 – K Exam 3 – Hamrick – (54868) 2 1. f − 1 ( x ) = parenleftBig x − 3 5 parenrightBig 1 / 3 2. f − 1 ( x ) = parenleftBig 3 − x 5 parenrightBig 1 / 3 3. f − 1 ( x ) = parenleftBig x − 5 3 parenrightBig 1 / 3 4. f − 1 ( x ) = parenleftBig 5 − x 3 parenrightBig 1 / 3 correct 5. f − 1 ( x ) = parenleftBig 3 + x 5 parenrightBig 1 / 3 6. f − 1 ( x ) = parenleftBig 5 + x 3 parenrightBig 1 / 3 Explanation: To determine f − 1 we solve for x in y = 5 − 3 x 3 and then interchange x, y . In this case 3 x 3 = 5 − y, i.e. x = parenleftBig 5 − y 3 parenrightBig 1 / 3 . Consequently, f − 1 ( x ) = parenleftBig 5 − x 3 parenrightBig 1 / 3 . 004 10.0 points Find the inverse, f − 1 , of f ( x ) = 1 − e x . 1. f − 1 ( x ) = ln parenleftBig 1 x − 1 parenrightBig , x > 1 2. f − 1 ( x ) = ln( x − 1) 3. f − 1 ( x ) = ln( x − 1) , x > 1 4. f − 1 ( x ) = ln(1 − x ) , x < 1 correct 5. f − 1 ( x ) = ln(1 − x ) 6. f − 1 ( x ) = ln parenleftBig 1 1 − x parenrightBig , x < 1 7. f − 1 ( x ) = ln parenleftBig 1 1 − x parenrightBig 8. f − 1 ( x ) = ln...
View Full Document

{[ snackBarMessage ]}

### Page1 / 8

K Exam 3-solutions - Version 037 – K Exam 3 – Hamrick...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online