{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Final Exam

# Final Exam - Garcia Ilse Final 1 Due 2:00 am Inst Fonken...

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

Garcia, Ilse – Final 1 – Due: Dec 18 2007, 2:00 am – Inst: Fonken 1 This print-out should have 25 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points At which point on the graph A B C D E F is the slope greatest ( i.e. , most positive)? 1. B 2. C correct 3. A 4. E 5. D 6. F Explanation: By inspection the point is C . keywords: slope, graph, change of slope 002 (part 1 of 1) 10 points Below is the graph of a function f . -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 2 4 - 2 - 4 2 4 - 2 - 4 Use the graph to determine lim x 3 f ( x ). 1. does not exist 2. limit = - 1 3. limit = - 3 correct 4. limit = 0 5. limit = - 2 Explanation: From the graph it is clear that the limit lim x 3 - f ( x ) = - 3 , from the left and the limit lim x 3+ f ( x ) = - 3 , from the right exist and coincide in value. Thus the two-sided lim x 3 f ( x ) = - 3 . keywords: limit, graph, limit at removable discontinuity 003 (part 1 of 1) 10 points Determine the limit lim x 6 8 (6 - x ) 2 . 1. none of the other answers

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

View Full Document
Garcia, Ilse – Final 1 – Due: Dec 18 2007, 2:00 am – Inst: Fonken 2 2. limit = correct 3. limit = 4 3 4. limit = - 4 3 5. limit = -∞ Explanation: Since lim x 0 ( x - 6) 2 = 0 and (6 - x ) 2 > 0 for all x 6 = 0, we see that lim x 6 8 (6 - x ) 2 = . keywords: limit, rational function 004 (part 1 of 1) 10 points Determine if lim x → ∞ x p x 2 + 4 - x · exists, and if it does, find its value. 1. limit = 1 2. limit = 3 3. limit = 2 correct 4. limit = 5 2 5. limit does not exist 6. limit = 3 2 Explanation: After rationalization we see that x p x 2 + 4 - x · = x x 2 + 4 - x 2 x 2 + 4 + x = 4 x x 2 + 4 + x = 4 r 1 + 4 x 2 + 1 . On the other hand, lim x → ∞ r 1 + 4 x 2 = 1 . Consequently, lim x → ∞ x p x 2 + 4 - x · exists and has limit = 2 . keywords: limit, limit at infinity, square root function, rationalize numerator 005 (part 1 of 1) 10 points Find the value of lim x 0 e 3 x - e - 3 x sin 2 x . 1. limit = 7 2 2. limit does not exist 3. limit = 3 2 4. limit = 3 correct 5. limit = 4 6. limit = 2 Explanation: Set f ( x ) = e 3 x - e - 3 x , g ( x ) = sin 2 x. Then f, g are everywhere differentiable func- tions such that lim x 0 f ( x ) = lim x 0 g ( x ) = 0 . Thus L’Hospital’s Rule applies: lim x 0 f ( x ) g ( x ) = lim x 0 f 0 ( x ) g 0 ( x ) .
Garcia, Ilse – Final 1 – Due: Dec 18 2007, 2:00 am – Inst: Fonken 3 Now f 0 ( x ) = 3( e 3 x + e - 3 x ) , g 0 ( x ) = 2 cos 2 x, while lim x 0 f 0 ( x ) = 6 , lim x 0 g 0 ( x ) = 2 . Consequently, lim x 0 e 3 x - e - 3 x sin 2 x = 3 . keywords: 006 (part 1 of 1) 10 points Find all the values of x at which the func- tion f defined by f ( x ) = x - 4 2 x 2 - 4 x - 16 is not continuous. 1. x = 2 , - 4 , - 2 2. no values of x 3. x = - 2 4. x = - 4 , 2 5. x = 4 , 2 6. x = 4 , - 2 correct Explanation: After factorization the denominator be- comes 2 x 2 - 4 x + 16 = 2( x - 4)( x + 2) , so f can be written as ( ) f ( x ) = x - 4 2 x 2 - 4 x + 16 = 1 2( x + 2) , whenever x 6 = 4. But for x = 4 the numerator and denominator are both zero. Thus f is not defined, hence not continuous, at x = 4. For other values of x the function f is rational, and so will fail to be continuous only at zeros of the denominator in ( ), i.e. at x = - 2.

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.

{[ snackBarMessage ]}