HW13-solutions

# HW13-solutions - white(taw933 – HW13 – ben-zvi...

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Unformatted text preview: white (taw933) – HW13 – ben-zvi – (55600) 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Determine which, if any, of f ( x ) = 256(16 − 3 x ) , g ( x ) = parenleftbigg 1 16 parenrightbigg 3 x − 2 , h ( x ) = 256(4 − 6 x ) , define the same function. 1. none of f, g, or h 2. only g, f 3. f, g, and h correct 4. only f, h 5. only g, h Explanation: By the laws of Exponents f ( x ) = 256(16 − 3 x ) = 16 2 (16 − 3 x ) = 16 2 − 3 x , while g ( x ) = parenleftbigg 1 16 parenrightbigg 3 x − 2 = 16 − (3 x − 2) = 16 2 − 3 x and h ( x ) = 256(4 − 6 x ) = 16 2 ( 4 2 ) − 3 x = 16 2 (16 − 3 x ) = 16 2 − 3 x Consequently, all of f, g, and h define the same function. 002 10.0 points Determine if lim x →−∞ parenleftbigg e 2 x + 2 e − 2 x 3 e 2 x + 5 e − 2 x parenrightbigg exists, and if it does, find its value. 1. limit = 2 5 correct 2. limit = 0 3. limit = 2 3 4. limit = 1 3 5. limit does not exist 6. limit = 1 5 Explanation: Since lim x →−∞ e − ax = ∞ , lim x →−∞ e ax = 0 when a > 0, evaluating the limit directly gives lim x →−∞ parenleftbigg e − 2 x + 2 e 2 x 3 e 2 x + 5 e − 2 x parenrightbigg = ∞ ∞ , which doesn’t make any sense. (And we can’t just cancel the ∞ ’s because infinities don’t work like that.) So we try to get rid of terms that go to ∞ and leave terms that go to zero. To achieve this, multiply top and bottom by e 2 x . Then e 2 x e 2 x parenleftbigg e 2 x + 2 e − 2 x 3 e 2 x + 5 e − 2 x parenrightbigg = e 4 x + 2 3 e 4 x + 5 , and so lim x →−∞ e − 2 x + 2 e 2 x 3 e 2 x + 5 e − 2 x = lim x →−∞ e 4 x + 2 3 e 4 x + 5 . white (taw933) – HW13 – ben-zvi – (55600) 2 But, as we noted, lim x →−∞ e 4 x = 0 , so by properties of limits, lim x →−∞ ( e 4 x + 2) = 2 , lim x →−∞ (3 e 4 x + 5) = 5 . By properties of limits again, therefore, lim x →−∞ parenleftbigg e 2 x + 2 e − 2 x 3 e 2 x + 5 e − 2 x parenrightbigg exists and the limit = 2 5 . keywords: exponential function, limit at in- finity, properties of limits, 003 10.0 points Find the derivative of f ( x ) = ( x 2 + x- 5) e − x . 1. f ′ ( x ) = (6- x + x 2 ) e − x 2. f ′ ( x ) = (6 + x- x 2 ) e − x correct 3. f ′ ( x ) = (6- x- x 2 ) e − x 4. f ′ ( x ) = (5 + x- x 2 ) e − x 5. f ′ ( x ) = (5- x + x 2 ) e − x 6. f ′ ( x ) = (5 + x + x 2 ) e − x Explanation: By the Product Rule, f ′ ( x ) = (2 x + 1) e − x- ( x 2 + x- 5) e − x ....
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HW13-solutions - white(taw933 – HW13 – ben-zvi...

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