{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

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

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

View Full Document Right Arrow Icon
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
Background image of page 1

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

View Full Document Right Arrow Icon
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 ) .
Background image of page 2
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.
Background image of page 3

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

View Full Document Right Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}