Curve sketching Problem (1 Viewer)

[Bladez]

Ok with: ln(x) + 1/x

is 1/x an asymptote as x -> 0? I suggested it to the teacher in class and all he could say was "I'm not sure that it is" so I'm a bit confused...

Also if you suggested there was an asymptote like this in a test that is kind of dodgy like this one (maybe... excuse my ramblings if they don't make sens) would you lose marks?

 

[shkspeare]

yes it is how can ur teacher not know

as x approaches 0+ y approaches inifinity

x approaches 0-, y approahces -infintiy

 

[George W. Bush]

x=0 is the asymptote, and shkspeare is wrong on both counts
as x approaches 0+, y approaches -infinity
the function is only defined for x>0 so there's no x approaching 0-

 

[shkspeare]

i thought he was talking about y=1/x sorry

 

[CM_Tutor]

Originally posted by George W. Bush
x=0 is the asymptote, and shkspeare is wrong on both counts
as x approaches 0+, y approaches -infinity
the function is only defined for x>0 so there's no x approaching 0-

Sorry, but as x --> 0⁺ y --> + inf
There are a couple of ways to show this:
1. For a function y = ln(x) + 1 / x, the polynomial will dominate the logarithm, and so the behaviour of 1 / x is dominant.
2. Consider dy/dx = (1 / x) - (1 / x²) = (x - 1) / x² - clearly we have a stationary point at (1, 1)
d²y/dx² = (-1 / x²) - (-2 / x³) = (2 - x) / x³. At (1, 1), d²y/dx² = 1 > 0, and thus our stationary point is, in fact, a minimum turning point.
3. As x --> 0⁺, dy/dx --> - inf, and so near x = 0, the function is decreasing at an extremely rapid rate.

x = 0 is certainly a vertical asymptote. As for the original question, is 1 / x an asymptote? I would say no, as the term ln(x) does not approach zero, it approaches - inf.

 

[George W. Bush]

It was a....errrrrrr......typo....

 

[Grey Council]

I was thinking of a dodgy way to say Bush sucks in geeky graph sketching terms, but i've decided to come out and say it.

George Bush? You suck.
^____^

relax, i'm joking, in case you haven't realised. lol

no, seriously:

1. For a function y = ln(x) + 1 / x, the polynomial will dominate the logarithm, and so the behaviour of 1 / x is dominant.
3. As x --> 0+, dy/dx --> - inf, and so near x = 0, the function is decreasing at an extremely rapid rate.

If the polynomial dominates, why would the whole graph:
As x --> 0+, dy/dx --> - inf
?!?!
1/x is near the top, as x->0, the whole graph will approach +inf, not -inf.

EDIT:
whoops, you said dy/dx --> -inf
my bad, didn't read properly.
hrmph, it IS 6:45am, lol.

 

[George W. Bush]

I'm sure you can write George Bush Sucks on a cartesian plane using careful restriction of domain and range