# bijection of e^x * ln x

• Jan 20th 2010, 11:50 AM
Gok2
bijection of e^x * ln x
Hey people.
I need to prove that f(x)=e^x * ln x gets every value in R exactly once. I managed to prove that f(x) gets every value in R , i.e. , for each y in R there is x in (0,inf) such as f(x)=y . But when I tried to show that f gets every value in R exactly once , i.e. showing that f is bijective , I kinda got stuck...

I tried to assume that there are x_1 , x_2 in (0,inf) such as f(x_1)=f(x_2).
f is differentiable in (0,inf) as a product of differentiable functions in (0,inf) , and therefore , from Rolle's theorem I can say that there is a point c in (x_1 , x_2) such as f'(c)=0 , i.e. , (e^x / x) + e^x*ln x = 0.
if x >= 1, then it is a contradiction , becuase for x>=1 , (e^x / x) + e^x*ln x is always positive. But if 0<x<1 , then e^x*ln x is negetive , and e^x / x is positive of course , so I can't say that f's derivative is always different from zero.. I tried to prove it without Rolle's theorem as well, but I didn't managed to... Any clues ?

Thanks!
• Jan 20th 2010, 12:22 PM
running-gag
Hi

You can simply show that f is strictly increasing over )0,+oo( and that its limit in 0 is -oo, and its limit in +oo is +oo
• Jan 20th 2010, 12:24 PM
Gok2
Hi,
I know how to show that f's limit at x->0^+ is -oo and at x->+oo is +oo , but how does that help me showing that f is strictly increasing in (0, inf) ?
• Jan 20th 2010, 12:28 PM
Jhevon
Quote:

Originally Posted by Gok2
Hi,
I know how to show that f's limit at x->0^+ is -oo and at x->+oo is +oo , but how does that help me showing that f is strictly increasing in (0, inf) ?

one way: show that its derivative is always positive on that interval.
• Jan 20th 2010, 12:31 PM
Plato
Quote:

Originally Posted by Gok2
Hi,
I know how to show that f's limit at x->0^+ is -oo and at x->+oo is +oo , but how does that help me showing that f is strictly increasing in (0, inf) ?

Have you graphed the derivative $\displaystyle e^x\ln(x)+\frac{e^x}{x}$ on $\displaystyle (0,2)$?
The graph will suggest some ways to proceed,
• Jan 20th 2010, 12:32 PM
Gok2
I know. f's derivative is always positive at the interval (1,inf), and therefore f is strictly increasing there.
But I don't know nothing about f's derivative at the interval (0,1) , becuase for each 0<x<1 , e^x /x is positive , and e^x*ln x is negative , so I don't really know much about f's derivative at the interval (0,1) , that's why I am stuck...
• Jan 20th 2010, 12:35 PM
Gok2
Quote:

Originally Posted by Plato
Have you graphed the derivative $\displaystyle e^x\ln(x)+\frac{e^x}{x}$ on $\displaystyle (0,2)$?
The graph will suggest some ways to proceed,

Yeah , f's derivative seems always to be positive , even at (0,1) . But I don't know how to show that algebraly .
• Jan 20th 2010, 12:38 PM
Drexel28
Quote:

Originally Posted by Gok2
Hey people.
I need to prove that f(x)=e^x * ln x gets every value in R exactly once. I managed to prove that f(x) gets every value in R , i.e. , for each y in R there is x in (0,inf) such as f(x)=y . But when I tried to show that f gets every value in R exactly once , i.e. showing that f is bijective , I kinda got stuck...

I tried to assume that there are x_1 , x_2 in (0,inf) such as f(x_1)=f(x_2).
f is differentiable in (0,inf) as a product of differentiable functions in (0,inf) , and therefore , from Rolle's theorem I can say that there is a point c in (x_1 , x_2) such as f'(c)=0 , i.e. , (e^x / x) + e^x*ln x = 0.
if x >= 1, then it is a contradiction , becuase for x>=1 , (e^x / x) + e^x*ln x is always positive. But if 0<x<1 , then e^x*ln x is negetive , and e^x / x is positive of course , so I can't say that f's derivative is always different from zero.. I tried to prove it without Rolle's theorem as well, but I didn't managed to... Any clues ?

Thanks!

Let $\displaystyle x,y\in\mathbb{R}$ be arbitrary. Assume WLOG that $\displaystyle x<y$ then $\displaystyle \ln(x)<\ln(y)$ and $\displaystyle 0<y-x\implies 1<e^{y-x}$ so that $\displaystyle \ln(x)<\ln(y)<\ln(y)e^{y-x}$ and so $\displaystyle \ln(x)e^x<\ln(y)e^y$. Thus, $\displaystyle x\ne y\implies f(x)\ne f(y)$ and so your function is injective.
• Jan 20th 2010, 12:46 PM
Gok2
Quote:

Originally Posted by Drexel28
Let $\displaystyle x,y\in\mathbb{R}$ be arbitrary. Assume WLOG that $\displaystyle x<y$ then $\displaystyle \ln(x)<\ln(y)$ and $\displaystyle 0<y-x\implies 1<e^{y-x}$ so that $\displaystyle \ln(x)<\ln(y)<\ln(y)e^{y-x}$ and so $\displaystyle \ln(x)e^x<\ln(y)e^y$. Thus, $\displaystyle x\ne y\implies f(x)\ne f(y)$ and so your function is injective.

Cool, didn't think about that . Thanks alot to you and everyone who tried to help!