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Unformatted text preview: Create assignment, 57321, Homework 2, Jan 30 at 9:14 am 1 This printout should have 28 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. The due time is Central time. CalC3a01a 49:01, calculus3, multiple choice, > 1 min, fixed. 001 If f is a differentiable function, then f ( a ) is given by which of the following? I. lim h → f ( a + h ) f ( a ) h II. lim x → a f ( x ) f ( a ) x a III. lim x → a f ( x + h ) f ( x ) h 1. I only 2. II only 3. I and II only correct 4. I and III only 5. I, II, and III Explanation: Both of f ( a ) = lim h → f ( a + h ) f ( a ) h and f ( a ) = lim x → a f ( x ) f ( a ) x a are valid definitions of f ( a ). By contrast, lim x → a f ( x + h ) f ( x ) h = f ( a + h ) f ( a ) h because f is continuous. Consequently, f ( a ) is given only by I and II . keywords: CalC3a01s 49:01, calculus3, multiple choice, > 1 min, wordingvariable. 002 What is the significance of the expression f (1 + h ) f (1) h in the following graph of f when h = 9 2 ? 1 2 3 4 5 1 2 3 4 5 P Q R S T U 1. slope of line through P and U correct 2. equation of line through P and U 3. length of line segment PU 4. slope of line through P and R 5. equation of line through P and R 6. length of line segment PR 7. slope of line through P and T 8. equation of line through P and T 9. length of line segment PT Create assignment, 57321, Homework 2, Jan 30 at 9:14 am 2 10. slope of tangent line at P Explanation: When h = 9 2 the expression f (1 + h ) f (1) h is the ratio of the rise and the run between the points P and U . Thus the expression is the slope of line through P and U . keywords: slope, secant line CalC3a02d 49:01, calculus3, multiple choice, < 1 min, normal. 003 Let f be the function defined by f ( x ) = 5 x + ( x 3 +  x 3  ) 2 . Determine if lim h → f (3 + h ) f (3) h exists, and if it does, find its value. 1. limit = 5 correct 2. limit = 8 3. limit = 7 4. limit = 6 5. limit doesn’t exist 6. limit = 4 Explanation: In computing lim h → f (3 + h ) f (3) h we can’t simply differentiate because the func tion is defined by a different formula f ( x ) = ‰ 5 x, x < 3, 5 x + 4( x 3) 2 , x ≥ 3 on each side of the limit point x = 3. We have to use onesided limits lim h → 0+ f (3 + h ) f (3) h and lim h → f (3 + h ) f (3) h . Now for h > 0, f (3 + h ) f (3) h = 5 h + 4 h 2 h , and so lim h → 0+ f (3 + h ) f (3) h = lim h → 0+ 5+4 h = 5 . On the other hand, for h < 0, f (3 + h ) f (3) h = 5 h, and so lim h → f (3 + h ) f (3) h = lim h → 0+ 5 = 5 . As the left and right onesided limits exist and coincide, the twosided limit exists and lim h → f (3 + h ) f (3) h = 5 ....
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This note was uploaded on 03/20/2008 for the course M 408c taught by Professor Mcadam during the Spring '06 term at University of Texas.
 Spring '06
 McAdam

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