how can I prove that the function $f(x)=e^x\ln x$
attains every real number as its value exactly once. (by proving that it's a monotonic continuous function)?
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Prove it bijective using the general mean value theorem. i.e: Prove it to be strictly monotonic conitnious function and evaluate the image of R*+ – Oussama Boussif Jun 19 '16 at 18:16
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Where is $f$ defined ? – Dietrich Burde Jun 19 '16 at 18:19
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2I see, yesterday was $e^x + \log x,$ today is $e^x \log x$ http://math.stackexchange.com/questions/1831147/prove-that-fx-ex-ln-x-attains-every-real-number-as-its-value-exactly – Will Jagy Jun 19 '16 at 18:42
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@WillJagy yes I had a mistake in my question and didn't saw it until all of those ppl answered.. so I decided to ask a new one. – D_R Jun 19 '16 at 19:21
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The function is defined for $x>0$; it should be easy to prove that $$ \lim_{x\to0^+}f(x)=-\infty,\qquad \lim_{x\to\infty}f(x)=\infty $$ so the range of the function is $\mathbb{R}$ because…
The derivative is $$ f'(x)=\frac{e^x}{x}(1+x\ln x) $$
Consider $g(x)=1+x\ln x$; then $$ g'(x)=1+\ln x $$ that vanishes for $x=e^{-1}$ and $g(e^{-1})=1-e^{-1}>0$.
Can you conclude?
egreg
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Take it's derivative: $f'(x)=e^x \ln x + \frac{e^x}{x}$ and note that it is always strictly greater than 0.
This shows that it is strictly increasing thus injective, so if every real number is reached, than also exactly once.
Now because $\lim_{x\rightarrow 0} f(x) = -\infty$ and $\lim_{x\rightarrow \infty} f(x) = \infty$, every real number is reached because of the mean value theorem.
Jens Renders
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