# Decreasing/concave upward: xlnx

• Apr 21st 2010, 04:02 AM
TsAmE
Decreasing/concave upward: xlnx
a) On what interval is f(x) = xlnx decreasing?
b) On what interval is f concave upward?

My attempt:

a) f'(x) = (x)d/dx(lnx) + (lnx)d/x(x)
= x * 1 / x + lnx
= lnx

According to lnx graph, f(x) decreasing when 0 < x < 1

b) f''(x) = 1 / x

According to 1 / x graph f concave upward when x < 0
• Apr 21st 2010, 04:06 AM
HallsofIvy
Quote:

Originally Posted by TsAmE
a) On what interval is f(x) = xlnx decreasing?
b) On what interval is f concave upward?

My attempt:

a) f'(x) = (x)d/dx(lnx) + (lnx)d/x(x)
= x * 1 / x + lnx
= lnx

x*(1/x)= 1, not 0!

Quote:

According to lnx graph, f(x) decreasing when 0 < x < 1

b) f''(x) = 1 / x

According to 1 / x graph f concave upward when x < 0
The derivative of x ln(x) is 1+ ln(x), not ln(x), but the second derivative is 1/x so this is correct.

For what values of x is 1+ ln(x)< 0?
• Apr 21st 2010, 04:28 AM
chisigma
Quote:

Originally Posted by TsAmE
b) On what interval is $\displaystyle f(x)= x\cdot \ln x$ concave upward?

My attempt:

b) f''(x) = 1 / x

According to 1 / x graph f concave upward when x < 0

There is only one minor problem: for $\displaystyle x<0$ the function $\displaystyle f(x) = x\cdot \ln x$ is not a real function, so that the concept of 'concave upward function' hasn't the usual meaning...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
• Apr 21st 2010, 04:35 AM
TsAmE
Quote:

Originally Posted by HallsofIvy
For what values of x is 1+ ln(x)< 0?

x + ln(x) < 0 when x < 1/e, ok that makes sense, and how would you get the 0 in (0, 1/e)? I got it for my wrong answer by sketching the lnx graph and looking at it, but clearly that is wrong in this case. For b I tried making an equality 1 / x > 0 to solve for x and find when it is concave upward, but that doesnt work.
• Apr 21st 2010, 12:06 PM
chisigma
In order to find a function 'concave upward' for $\displaystyle x<0$ we can start to the second order DE...

$\displaystyle y^{''} = \frac{1}{x}$ (1)

The (1) is easily integrable and is...

$\displaystyle y(x) = x\cdot \ln |x| + c_{1}\cdot x + c_{2}$ (2)

If we impose the conditions $\displaystyle y(0)=0$ and $\displaystyle y^{'} (1)=1$ we obtain $\displaystyle c_{1}= c_{2} =0$ so that the function we are loocking for is...

$\displaystyle y(x)= x\cdot \ln |x|$ (3)

... a 'beautiful odd function'...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
• Apr 21st 2010, 01:53 PM
TsAmE
Sorry but I havent done integration yet, only doing that later this year
• Apr 22nd 2010, 06:09 AM
chisigma
The 'beautiful odd function' $\displaystyle f(x)= x\cdot \ln |x|$ is illustrate there...

http://digilander.libero.it/luposabatini/MHF56.bmp

It is a 'upward concave function' for $\displaystyle x<0$ , whereas $\displaystyle x\cdot \ln x$ has complex value in that interval...

The reason for which I have called this function 'beautiful' is relative to a ' still unresolved problem' the solution of which is strictly connected to it...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
• Apr 23rd 2010, 03:06 AM
ione
Quote:

Originally Posted by chisigma
The 'beautiful odd function' $\displaystyle f(x)= x\cdot \ln |x|$ is illustrate there...

http://digilander.libero.it/luposabatini/MHF56.bmp

It is a 'upward concave function' for $\displaystyle x<0$ , whereas $\displaystyle x\cdot \ln x$ has complex value in that interval...

The reason for which I have called this function 'beautiful' is relative to a ' still unresolved problem' the solution of which is strictly connected to it...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

Isn't it an 'upward concave function' for $\displaystyle x>0$?
• Apr 23rd 2010, 05:38 AM
chisigma
According to...

Convex function - Wikipedia, the free encyclopedia

... is seems that ione is right!... on effect 'upward concave' is the same as 'convex' , so that $\displaystyle f(x)= x\cdot \ln |x|$ is convex for $\displaystyle x>0$ and concave for $\displaystyle x<0$ ...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$