{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

HW5-solutions

# HW5-solutions - wei(jw35975 HW5 milburn(54685 This...

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

wei (jw35975) – HW5 – milburn – (54685) 1 This print-out should have 25 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points Determine lim x 1 b 1 x 2 x 1 x 1 B . 1. limit = 1 correct 2. limit = 1 2 3. limit does not exist 4. limit = 1 3 5. limit = 1 6. limit = 1 3 7. limit = 1 2 Explanation: AFter simplifcation we see that 1 x 2 x 1 x 1 = 1 x x ( x 1) = 1 x For all x n = 1. Thus limit = lim x 1 1 x = 1 . 002 (part 1 of 3) 10.0 points Let F be the Function defned by F ( x ) = x 2 64 | x 8 | . (i) Determine lim x 8 + F ( x ) . 1. limit = 16 2. limit does not exist 3. limit = 8 4. limit = 8 5. limit = 16 correct Explanation: AFter Factorization, x 2 64 | x 8 | = ( x + 8)( x 8) | x 8 | . But, For x > 8, | x 8 | = x 8 . Thus F ( x ) = x + 8 , x > 8 , in which case, by properties oF limits, the right hand limit lim x 8 + F ( x ) = 16 . 003 (part 2 of 3) 10.0 points (ii) Determine lim x 8 - F ( x ) . 1. limit does not exist 2. limit = 16 correct 3. limit = 8 4. limit = 16 5. limit = 8 Explanation: AFter Factorization, x 2 64 | x 8 | = ( x + 8)( x 8) | x 8 | . But, For x < 8, | x 8 | = ( x 8) .

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

View Full Document
wei (jw35975) – HW5 – milburn – (54685) 2 Thus F ( x ) = ( x + 8) , x < 8 , in which case, by properties of limits, the left hand limit lim x 8 - F ( x ) = 16 . 004 (part 3 of 3) 10.0 points (iii) Use your results for parts (i) and (ii) to determine lim x 8 F ( x ) . 1. limit = 16 2. limit = 8 3. limit = 8 4. limit = 16 5. limit does not exist correct Explanation: By parts (i) and (ii), lim x 8 + F ( x ) n = lim x 8 - F ( x ) . Consequently, the two-sided limit does not exist . 005 10.0 points Determine lim x 1 x 1 x + 3 2 . 1. limit = 2 2. limit = 4 correct 3. limit = 1 4 4. limit doesn’t exist 5. limit = 1 2 Explanation: After rationalizing the denominator we see that 1 x + 3 2 = x + 3 + 2 ( x + 3) 4 = x + 3 + 2 x 1 . Thus x 1 x + 3 2 = x + 3 + 2 for all x n = 1. Consequently, limit = lim x 1 ( x + 3 + 2) = 4 . 006 10.0 points Determine if lim h 0 f (1 + h ) f (1) h exists when f ( x ) = x 2 + 5 x , and if it does, Fnd its value. 1. limit = 7 correct 2. limit = 11 3. limit does not exist 4. limit = 8 5. limit = 10 6. limit = 9 Explanation: Since f (1 + h ) f (1) = (6 + 7 h + h 2 ) 6 , we see that f (1 + h ) f (1) h = h 2 + 7 h h = h + 7 .
wei (jw35975) – HW5 – milburn – (54685) 3 On the other hand, lim h 0 ( h + 7) = 7 , by Properties of Limits. Consequently, the given limit exists, and limit = 7 . keywords: limit, deFnition of derivative 007 10.0 points ±unctions f and g are deFned on ( 10 , 10) by their respective graphs in 2 4 6 8 2 4 6 8 4 8 4 8 f g ±ind all values of x where the product, fg , of f and g is continuous, expressing your answer in interval notation. 1. ( 10 , 2) u ( 2 , 10) 2. ( 10 , 10) correct 3. ( 10 , 2) u ( 2 , 4) u (4 , 10) 4. ( 10 , 2] u [4 , 10) 5. ( 10 , 4) u (4 , 10) Explanation: Since f and g are piecewise linear, they are continuous individually on ( 10 , 10) except at their ‘jumps’; i.e. , at x = 4 in the case of f and x = 4 , 2 in the case of g . But the product of continuous functions is again continuous, so is certainly continuous on ( 10 , 2) u ( 2 , 4) u (4 , 10) .

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 ]}

### Page1 / 11

HW5-solutions - wei(jw35975 HW5 milburn(54685 This...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online