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# bonusAsol - Solutions to Bonus Quiz A...

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Unformatted text preview: Solutions to Bonus Quiz A www.math.uﬂ.edu/˜harringt February 17, 2008 1. Find all the vertical asymptotes for f (x) = Note that f (x) = (x−1) sin(x) x3 − x = x− 1 x− 1 ∗ sin x x ∗ (x−1) sin(x) . x3 − x 1 . x+1 So we have holes at x = 1 and at x = 0 since limx→0 limx→1 x−1 = 1. x− 1 sin x x = 1 and Thus, we have ONE vertical asymptote at x = −1. 2. Solve for x: e3x = 5e4x+2 Note: There are multiple ways to do this problem. I will demonstrate only two. e3x = e3x − 5e4x+2 = e3x − 5e3x+x+2 = e3x (1 − 5ex+2 ) = 1 − 5ex+2 = 1= 1/5 = ln 1/5 − 2 = 5e4x+2 0 0 0 Note: that e3x = 0 for all x 0 We divided both sides by e3x 5ex+2 ex+2 x Or.... 1 e3x ln(e3 x) 3x 3x −x x 5e4x+2 ln(5 ∗ e4x+2 ) ln(5) + ln(e4x+2 ) ln(5) + 4x + 2 ln(5) + 2 − ln(5) − 2 = = = = = = 3. Evaluate the following limit: limx→0 tan x x tan x sin(x) sin(x) 1 = lim = lim ∗ lim =1∗1=1 x→0 x→0 x cos(x) x→0 x→0 cos x x x lim 4. Determine which of the following is even: x sin x or answer. |x| . x Explain your Recall that a function is even when f (−x) = f (x). (−x) sin(−x) = −x ∗ (− sin(x)) = x ∗ sin x Thus, x sin x is EVEN. |−x| −x = |x| −x = − |x| Hence, x |x| x is ODD. 5. Compute the following limits. Let 3 x + 1 for x < 0 x2 − x for 0 ≤ x < 1 f (x) = √ x + x for x > 1 (a) limx→0 f (x) =Does Not Exist Since limx→0+ f (x) = 02 + 0 = 0 and limx→0− f (x) = 03 + 1 = 1 (b) limx→1− f (x) = 12 + 1 = 2 (c) limx→0+ f (x) = 02 + 0 = 0 (d) limx→1 f (x) = 2 √ Since limx→1+ f (x) = 1 + 1 = 2 and limx→1− f (x) = 12 + 1 = 2 2 ...
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## This note was uploaded on 07/10/2011 for the course MAC 2311 taught by Professor All during the Fall '08 term at University of Florida.

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bonusAsol - Solutions to Bonus Quiz A...

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