# Weird natural log question

• August 21st 2008, 05:36 PM
BogStandard
Weird natural log question
You are given the function

f(x) = x*lnx

(x > 0)

a) Find the solution(s) of the equation f(x) = 0
b) Find any minima/maxima that f(x) has.

c) Find the value(s) of x0 for which

INTEGRAL ( between limits x0 and 0) f(x) dx = 0

and describe graphically what this means.

Ok first of all, for part a) all I can think of for a solution is x = 1 because lnx = 0 only when x = 1, right? Thing is this question is worth 4 marks so I feel like im missing something.

Secondly for part b) I didnt think there were any maxima/minima because of what I said above. However, differentiating the function I get

lnx + x(1/x)
equating to zero gives

lnx + 1 = 0
lnx = -1
lnx = -lne
lnx = lne^-1

x = 1/e

Plugging this into the second derivative (which is 1/x) gets me e as a minima.
Is this correct?

And for the last part it tells me to use integration by parts, but it doesnt get me anywhere...
• August 21st 2008, 05:51 PM
Jhevon
Quote:

Originally Posted by BogStandard
You are given the function

f(x) = x*lnx

(x > 0)

a) Find the solution(s) of the equation f(x) = 0
b) Find any minima/maxima that f(x) has.

c) Find the value(s) of x0 for which

INTEGRAL ( between limits x0 and 0) f(x) dx = 0

and describe graphically what this means.

Ok first of all, for part a) all I can think of for a solution is x = 1 because lnx = 0 only when x = 1, right? Thing is this question is worth 4 marks so I feel like im missing something.

if x*ln(x) = 0 then either x = 0 or ln(x) = 0 because if you have a product of two terms that yield zero, one or the other must be zero. however, we have that x > 0 (we need this for the log to be defined) so, you are correct, you are left with only one solution, x = 1.

Quote:

Secondly for part b) I didnt think there were any maxima/minima because of what I said above. However, differentiating the function I get

lnx + x(1/x)
equating to zero gives

lnx + 1 = 0
lnx = -1
lnx = -lne
lnx = lne^-1

x = 1/e
good!

Quote:

Plugging this into the second derivative (which is 1/x) gets me e as a minima.
Is this correct?
yup. you're a genius!

Quote:

And for the last part it tells me to use integration by parts, but it doesnt get me anywhere...
use integration by parts with $u = \ln x$ and $dv = x$

you know what u and dv represent, right?