How do you draw (lnx)/x?

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 :smile:

Reply 1

TeeEm

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 :smile:

look at Q21 in this link

http://madasmaths.com/archive/maths_booklets/further_topics/various/curve_sketching.pdf

Reply 2

AK 12

Consider transformation of

1/x then simply inx/x not hard mate.

Reply 3

Plagioclase

Original post by TeeEm

look at Q21 in this link

http://madasmaths.com/archive/maths_booklets/further_topics/various/curve_sketching.pdf

Sorry to hijack this question but do you have any idea why when I input this into WolframAlpha it plots values for

x \leq 0

? Is it assuming that

ln(x)=ln(|x|)

Reply 4

TeeEm

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

x \leq 0

? Is it assuming that

ln(x)=ln(|x|)

no idea
something to do with their settings...

lnx is only defined for x>0

Reply 5

Honeybadger32175

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 :smile:

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#nord=1&q=(ln+x)%2Fx

Reply 6

Zacken

First thing first, domain:

x > 0

, since the domain of ln x is x>0

Then look at the vertical asymptotes:

x=0

.

Then, horizontal aymptotes:

\lim_{x \to \infty} \frac{\ln x}{x} = 0

, so the y-axis is a horizontal asymptote.

Investigate

\lim_{x\to 0} \frac{\ln x}{x}

Spoiler

Roots:

\ln x = 0 \iff x = ??

Stationary points and their values there, use the quotient rule and solve

\displaystyle \left(\frac{\ln x}{x}\right)'=0

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.

Reply 7

Zacken

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

x \leq 0

? Is it assuming that

ln(x)=ln(|x|)

For

x < 0

, wolfram alpha plots the real part in blue and the imaginary part in orange. We assume that

f(x) = \frac{\ln x}{x}

is a real-valued function so it takes the domain

x > 0

, but it we work in the complex field then

f(z)

outputs complex numbers for

x < 0

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)

Reply 8

Plagioclase

Original post by Zacken

For

x < 0

, wolfram alpha plots the real part in blue and the imaginary part in orange. We assume that

f(x) = \frac{\ln x}{x}

is a real-valued function so it takes the domain

x > 0

, but it we work in the complex field then

f(z)

outputs complex numbers for

x < 0

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)