{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# calc9 - pokharel(yp624 HW09 Radin(56520 This print-out...

This preview shows pages 1–3. Sign up to view the full content.

pokharel (yp624) – HW09 – Radin – (56520) 1 This print-out should have 21 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Determine if the function f ( x ) = x x + 6 satisfies the hypotheses of Rolle’s Theorem on the interval [ 6 , 0], and if it does, find all numbers c satisfying the conclusion of that theorem. 1. c = 3 2. hypotheses not satisfied 3. c = 4 , 5 4. c = 4 , 4 5. c = 4 correct 6. c = 5 Explanation: Rolle’s Theorem says that if f is 1. continuous on [ a, b ] , 2. differentiable on ( a, b ) , and 3. f ( a ) = f ( b ) = 0, then there exists at least one c , a < c < b , such that f ( c ) = 0. Now the given function f ( x ) = x x + 6 , is defined for all x ≥ − 6, is continuous on [ 6 , ), and differentiable on ( 6 , ). In addition f ( 6) = f (0) = 0 . In particular, therefore, Rolle’s theorem ap- plies to f on [ 6 , 0]. On the other hand, by the Product and Chain Rules, f ( x ) = x + 6 + x 2 x + 6 = 3 x + 12 2 x + 6 . Thus there exists c, 6 < c < 0, such that f ( c ) = 3 c + 12 2 c + 6 = 0 , in which case c = 4 . 002 10.0 points Let f be a function defined on [0 , 1] such that f (0) = 1 , f (1) = 2 . Consider the following properties that f might have: A. f ( x ) = braceleftBigg x 2 + 1 , x negationslash = 1 / 2, 1 , x = 1 / 2 ; B. f ( x ) = | 3 x 1 | ; C. f is cont. on [0 , 1] , diff. on (0 , 1) ; which properties ensure that f ( c ) = 1 for some c in (0 , 1)? 1. all of them 2. A only 3. C only correct 4. A and B only 5. A and C only 6. B only 7. none of them 8. B and C only Explanation: The Mean Value Theorem (MVT) says:

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

View Full Document
pokharel (yp624) – HW09 – Radin – (56520) 2 If f is a function continuous on [ a, b ] and differentiable on ( a, b ) , then f ( c ) = f ( b ) f ( a ) b a for at least one c in ( a, b ) . If one or more of the hypotheses fail, then there need not exist any such c . A. No c : ( f ( x ) = 2 x, x negationslash = 1 / 2). B. No c : f ( x ) = braceleftBigg 3 , x < 1 / 3, 3 , x > 1 / 3. C. Does: (MVT). 003 10.0 points How many real roots does the equation x 5 + 4 x + 8 = 0 have? 1. exactly two real roots 2. exactly one real root correct 3. exactly four real roots 4. exactly three real roots 5. no real roots Explanation: Define f by f ( x ) = x 5 + 4 x + 8 . Then the roots of the equation x 5 + 4 x + 8 = 0 are the x -intercepts of the graph of f . Now f ( x ) x 5 for | x | large, so f ( x ) → ∞ as x → ∞ , while f ( x ) → −∞ as x → −∞ . Thus the graph of f must cross the x -axis at least once. Suppose the graph crosses the x at values x = a and x = b with a < b , i.e. , f ( a ) = f ( b ) = 0 , ( a < b ) . On the other hand, f is a polynomial function, it is continuous and differentiable for all x . Hence Rolle’s Theorem applies, so there exists some c , a < c < b , at which f ( c ) = 0. But f ( x ) = 5 x 4 + 4 > 0 for all x which isn’t consistent with f ( c ) = 0 for some c . Consequently, the graph of f can’t have x -intercepts at both x = a and x = b , so the equation has exactly one root .
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}