Exam3 12 - x = 0 and is given by Now the given Function is...

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Betancourt, Daniel – Exam 3 – Due: May 1 2007, 11:00 pm – Inst: Diane Radin 12 6. Explanation: Let’s frst review some properties oF ln x and ln( - x ). Since ln x is defned only on (0 , ) and lim x 0 + ln x = -∞ , lim x →∞ ln x = , the graph oF ln x has a vertical asymptote at x = 0 and so is given by But then ln( - x ) is defned only on ( -∞ , 0) and has the properties lim x 0 - ln( - x ) = -∞ , lim x →-∞ ln( - x ) = , so its graph has a vertical asymptote at
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Unformatted text preview: x = 0 and is given by Now the given Function is f ( x ) = log 3 ‡ 1 4-x · =-log 3 (4-x ) . Its graph will have a vertical asymptote at x = 4, and so will be that oF log 3 (-x ) translated 4 units to the leFt, then ‘±ipped over’ the x-axis. Consequently, f has graph keywords: graph, log Function, vertical aymp-tote, limit at infnity, absolute value Function...
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This note was uploaded on 02/14/2012 for the course MATH 408 K taught by Professor Clark,c.w./hoy,r.r during the Spring '08 term at University of Texas.

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