{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# exam 2 - Version 033 Exam 2 Radin(56520 This print-out...

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

Version 033 – Exam 2 – Radin – (56520) 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 Determine if lim x → ∞ x 3 4 x 5 x 3 + 4 x 2 + 5 exists, and if it does, find its value. 1. limit does not exist 2. limit = 0 3. limit = 1 5 4. limit = 2 5 5. limit = 3 5 6. limit = 1 5 correct Explanation: After division by x 3 in both the numerator and denominator, we see that x 3 4 x 5 x 3 4 x 2 + 5 = 1 4 x 2 5 + 4 x + 5 x 3 . On the other hand, lim x → ∞ parenleftBig 1 4 x 2 parenrightBig = 1 , while lim x → ∞ parenleftBig 5 + 4 x + 5 x 3 parenrightBig = 5 . Consequently, lim x → ∞ x 3 4 x 5 x 3 + 4 x 2 + 5 exists and limit = 1 5 . 002 10.0 points Find the rational function f so the figure below is the graph of y = f ( x ) on [ 8 , 8]. -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 2 4 6 2 4 6 2 4 6 2 4 6 1. f ( x ) = x 2 + 3 x + 8 x 2 4 2. f ( x ) = x 2 3 x + 8 x 2 4 3. f ( x ) = x 2 3 x + 8 4 x 2 4. f ( x ) = x 2 + 3 x 8 x 2 4 correct 5. f ( x ) = x 2 + 3 x 8 x 2 6. f ( x ) = x 2 + 3 x 8 x 2 + 4 Explanation: The graph has vertical asymptotes at x = ± 2, so a term x 2 4 must appear in the denominator of f ( x ); in particular, therefore, f ( x ) negationslash = x 2 + 3 x 8 x 2 + 4 and f ( x ) negationslash = x 2 + 3 x 8 x 2 ,

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

View Full Document
Version 033 – Exam 2 – Radin – (56520) 2 eliminating two of the possible answers. The graph also has a horizontal asymptote y = 1, so the ratio of coefficients of degree 2 terms in the numerator and denominator of f ( x ) must be 1. Thus f ( x ) negationslash = x 2 3 x + 8 4 x 2 , eliminating a third answer. Hence one of f ( x ) = x 2 + 3 x 8 x 2 4 or f ( x ) = x 2 3 x + 8 x 2 4 or f ( x ) = x 2 + 3 x + 8 x 2 4 must hold. But the y -intercept of the graph is positive, while the y -intercepts of these possi- ble f are f (0) = 2 , or 2 and only one has the property f (0) = 2. Con- sequently, f ( x ) = x 2 + 3 x 8 x 2 4 . 003 10.0 points Determine which of the following graphs could be that of f ( x ) = 3 + 2 x 2 1 + x 2 . 1. 2. 3. 4. correct Explanation: The line y = 2 is a horizontal asymptote because lim x → ±∞ 3 + 2 x 2 1 + x 2 = 2 . On the other hand, f ( x ) = 2 x (1 + x 2 ) 2 . Thus the critical point of f occurs at x = 0. In addition, f is increasing on ( −∞ , 0] and decreasing on [0 , ); in particular, therefore, f has a local maximum at x = 0. The only graph with these properties is
Version 033 – Exam 2 – Radin – (56520) 3 004 10.0 points If f is the function defined on [ 4 , 4] by f ( x ) = x + | x | − 4 , which of the following properties does f have? A. Absolute minimum at x = 0 . B. Differentiable at x = 0 . 1. B only 2. both of them 3. neither of them 4. A only correct Explanation: Since | x | = braceleftbigg x , x 0 , x , x < 0 , we see that f ( x ) = x + | x |− 4 = braceleftbigg 2 x 4 , x 0 , 4 , x < 0 . Thus on [ 4 , 4] the graph of f is 2 4 2 4 2 4 2 4 Consequently: A. True: by inspection.

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 ]}