Exam II Practice

# Exam II Practice - padilla(tp5647 – Review Exam 02 –...

This preview shows pages 1–4. Sign up to view the full content.

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

View Full Document

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: padilla (tp5647) – Review Exam 02 – Fouli – (58320) 1 This print-out should have 32 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points In the right-angled triangle θ 5 x the angle θ is increasing at a constant rate of 5 radians per hour. At what rate is the side of length x increas- ing when x = 3 feet? 1. rate = 26 ft/hour 2. rate = 20 ft/hour correct 3. rate = 24 ft/hour 4. rate = 22 ft/hour 5. rate = 28 ft/hour Explanation: By right-angle trigonometry, sin θ = x 5 , so cos θ dθ dt = 1 5 dx dt , in which case dx dt = 5 cos θ dθ dt = 25 cos θ . But by the properties of 3-4-5 right-angle tri- angles, cos θ = 4 / 5. Consequently, dx dt = 20 ft/hour . 002 10.0 points Which one of the following is the graph of f ( x ) = x x 2- 4 ? 1. 2 4- 2- 4 2 4- 2- 4 2. 2 4- 2- 4 2 4- 2- 4 correct 3. 2 4- 2- 4 2 4- 2- 4 4. 2 4- 2- 4 2 4- 2- 4 padilla (tp5647) – Review Exam 02 – Fouli – (58320) 2 5. 2 4- 2- 4 2 4- 2- 4 Explanation: The graph will have vertical asymptotes at x = ± 2 and horizontal asymptote at y = 0. On the other hand, f ′ ( x ) =- x 2 + 4 ( x 2- 4) 2 , so f will be decreasing on (-∞ ,- 2) , (- 2 , 2) , (2 , ∞ ) . The only graph having these is properties is 2 4- 2- 4 2 4- 2- 4 Consequently, this must be the graph of f . 003 10.0 points Determine the derivative of f ( x ) = x radicalbig x 2- 1 . 1. f ′ ( x ) = 2- x 2 √ x 2- 1 2. f ′ ( x ) = x 2- 2 √ x 2- 1 3. f ′ ( x ) = x 2 + 2 √ x 2- 1 4. f ′ ( x ) = 1- 2 x 2 √ x 2- 1 5. f ′ ( x ) = 2 x 2- 1 √ x 2- 1 correct 6. f ′ ( x ) = 2 x 2 + 1 √ x 2- 1 Explanation: By the Product and Chain Rules, f ′ ( x ) = radicalbig x 2- 1 + x 2 √ x 2- 1 = ( x 2- 1) + x 2 √ x 2- 1 . Consequently, f ′ ( x ) = 2 x 2- 1 √ x 2- 1 . keywords: 004 10.0 points The derivative, f ′ , of f has graph a b c graph of f ′ Use it to locate the critical point(s) x at which f has a local minimum? 1. x = b , c padilla (tp5647) – Review Exam 02 – Fouli – (58320) 3 2. x = a , b 3. x = b 4. x = c , a 5. x = a , b , c 6. x = c 7. x = a correct 8. none of a , b , c Explanation: Since the graph of f ′ ( x ) has no ‘holes’, the only critical points of f occur at the x- intercepts of the graph of f ′ , i.e. , at x = a, b, and c . Now by the first derivative test, f will have (i) a local maximum at x if f ′ ( x ) changes from positive to negative as x passes through x ; (ii) a local minimum at x if f ′ ( x ) changes from negative to positive as x passes through x . Consequently, by looking at the sign of f ′ ( x ) near each of x = a, b, and c we see that f has a local minimum only at x = a ....
View Full Document

## This note was uploaded on 09/09/2009 for the course CH 302 taught by Professor Mccord during the Spring '09 term at University of Texas-Tyler.

### Page1 / 17

Exam II Practice - padilla(tp5647 – Review Exam 02 –...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online