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# How do you draw (lnx)/x? watch

1. I know what the graph of lnx,1/x and the line y=x looks like but I have no idea how to put it together.

Any help is appreciated
2. (Original post by Super199)
I know what the graph of lnx,1/x and the line y=x looks like but I have no idea how to put it together.

Any help is appreciated
look at Q21 in this link

3. Consider transformation of

1/x then simply inx/x not hard mate.
4. Sorry to hijack this question but do you have any idea why when I input this into WolframAlpha it plots values for ? Is it assuming that ?
5. (Original post by Plagioclase)
Sorry to hijack this question but do you have any idea why when I input this into WolframAlpha it plots values for ? Is it assuming that ?
no idea
something to do with their settings...

lnx is only defined for x>0
6. (Original post by Super199)
I know what the graph of lnx,1/x and the line y=x looks like but I have no idea how to put it together.

Any help is appreciated
Not sure if this helps but if you put (ln x)/x into google it should come up with a graph. You can pick a point 'x' and it will also give you the 'y' value.

Or you can use this : http://www.google.co.uk/webhp?nord=1...1&q=(ln+x)%2Fx
7. First thing first, domain: , since the domain of ln x is x>0

Then look at the vertical asymptotes: .

Then, horizontal aymptotes: , so the y-axis is a horizontal asymptote.

Investigate
Spoiler:
Show
Can you see why it diverges to -infinity?

Roots:

Stationary points and their values there, use the quotient rule and solve and then determine whether the stationary point will be a maximum or a minimum by looking at the second derivative or (preferably in my opinion, investigating sign changes of the first derivative).

This should allow you to plot it all.
8. (Original post by Plagioclase)
Sorry to hijack this question but do you have any idea why when I input this into WolframAlpha it plots values for ? Is it assuming that ?
For , wolfram alpha plots the real part in blue and the imaginary part in orange. We assume that is a real-valued function so it takes the domain , but it we work in the complex field then outputs complex numbers for which WA attempts to plot by plotting the real and imaginary part separately. (To plot the actual curve in it's entirety would require four dimensions)
9. (Original post by Zacken)
For , wolfram alpha plots the real part in blue and the imaginary part in orange. We assume that is a real-valued function so it takes the domain , but it we work in the complex field then outputs complex numbers for which WA attempts to plot by plotting the real and imaginary part separately. (To plot the actual curve in it's entirety would require four dimensions)
Oh right, that makes sense. Thanks!
10. (Original post by Plagioclase)
Oh right, that makes sense. Thanks!
You're very welcome!
11. (Original post by AK 12)
Consider transformation of

1/x then simply inx/x not hard mate.
Wot mate? I can't possibly see how that's helpful.

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Updated: November 2, 2015
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