Thread: Fonction Help

1. Fonction Help

Hey friends

i got a Hard problem about Fonction:

We consider the function set to]0; + infinity[ g (x) = -lnx + 1 / x

A. Studying the variations of g
B. Show that the equation g (x) = 0, admits one and only one solution note (alfa) between 3/2 and 2

C. Consider the sign of g (x) sure everyone intervals]0;alfa[ and ]alfa; + infinity [

D. F is the function of R to R defined by f (x) = e ^ -x . lnx
what the sign of f '

sorry for my english (i am french educated ) , Just i need your Help to solve this ,

Thanks in advance

2. Originally Posted by iceman1
Hey friends

i got a Hard problem about Fonction:

We consider the function set to]0; + infinity[ g (x) = ln-x + 1 / x

A. Studying the variations of g
B. Show that the equation g (x) = 0, admits one and only one solution note (alfa) between 3/2 and 2

C. Consider the sign of g (x) sure everyone intervals]0;alfa[ and ]alfa; + infinity [

D. F is the function of R to R defined by f (x) = e ^ x-lnx
what the sign of f '

sorry for my english (i am french educated ) , Just i need your Help to solve this ,

Thanks in advance
g (x) = ln-x + 1 / x can be read in a number of ways - none that make sense within the context of the question. Please give the rule again - carefully formatted.

3. Originally Posted by mr fantastic
g (x) = ln-x + 1 / x can be read in a number of ways - none that make sense within the context of the question. Please give the rule again - carefully formatted.
g(x) = -Lnx + 1/x Like this==> ( -Lnx) + (1/x) good now?

4. Originally Posted by iceman1
g(x) = -Lnx + 1/x Like this==> ( -Lnx) + (1/x) good now?
Yes. - ln(x) is quite different to ln(-x).

Originally Posted by iceman1
[snip]B. Show that the equation g (x) = 0, admits one and only one solution note (alfa) between 3/2 and 2
[snip]
g(3/2) > 0 and g(2) < 0. Therefore there's at least one solution between 3/2 and 2. Assume there's more than one solution, x = a and x = b, say. Apply Rolle's Theorem. Get a contradiction. Therefore only one solution.