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f(x) = 1/x + ln(1+e^x)

2007-07-13 17:52:10 · 4 answers · asked by sahsjing 7 in Science & Mathematics Mathematics

4 answers

Ok, 1/x has x and y axes as asymptotes and is basically a couple of parabolas in the top right and bottom left quadrants of the plane.

For negative x, 1+e^x tends to 1 as x tends to infinity, so the log of it tends to zero and the bottom left parabola stays roughly the same.

But for positive x, 1+e^x is roughly e^x and its log is therefore roughly x, so we have x +1/x which looks like 1/x to start with for x near zero, but then the x part takes over as x tends to infinity, so its a parabola in the wedge between the positive y axis and the positive part of the line y=x.

So the asymptotes are the positive and negative y-axis, the negative x-axis and the positive part of the line y=x. That's three or four asymptotes depending on whether you count the positive and negative parts of the y-axis as separate or not. I would say probably not, so the answer is three.

2007-07-13 18:18:31 · answer #1 · answered by okei 4 · 2 0

i think/know there are 2 asymptotes: x = 0 and y = x

2016-01-17 08:24:26 · answer #2 · answered by Anonymous · 0 0

x = 0, y = 0, y = +x

2007-07-13 18:09:17 · answer #3 · answered by Helmut 7 · 0 0

I'd say three - one vertical, one horizontal and one sloping.

2007-07-13 18:11:01 · answer #4 · answered by Anonymous · 0 0

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