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Math Help - bijection of e^x * ln x

  1. #1
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    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!
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  2. #2
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    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
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  3. #3
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    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) ?
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Gok2 View Post
    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.
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  5. #5
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    Quote Originally Posted by Gok2 View Post
    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 e^x\ln(x)+\frac{e^x}{x} on (0,2)?
    The graph will suggest some ways to proceed,
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  6. #6
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    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...
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  7. #7
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    Quote Originally Posted by Plato View Post
    Have you graphed the derivative e^x\ln(x)+\frac{e^x}{x} on (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 .
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  8. #8
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Gok2 View Post
    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 x,y\in\mathbb{R} be arbitrary. Assume WLOG that x<y then \ln(x)<\ln(y) and 0<y-x\implies 1<e^{y-x} so that \ln(x)<\ln(y)<\ln(y)e^{y-x} and so \ln(x)e^x<\ln(y)e^y. Thus, x\ne y\implies f(x)\ne f(y) and so your function is injective.
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  9. #9
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    Quote Originally Posted by Drexel28 View Post
    Let x,y\in\mathbb{R} be arbitrary. Assume WLOG that x<y then \ln(x)<\ln(y) and 0<y-x\implies 1<e^{y-x} so that \ln(x)<\ln(y)<\ln(y)e^{y-x} and so \ln(x)e^x<\ln(y)e^y. Thus, 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!
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