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How to rigorously prove thar $h(x) = \log(1 + e^x)$ has minimum at $-\infty$?

I have tried taking the derivative of which comes $\frac{e^x}{1+e^x}=0\implies e^x=0$. This can only happen if $x\rightarrow -\infty$ also since it is a monotonically increasing function which is always +ve

I am not sure if this is a rigorous solution. Are there any counter-examples to this method? Any help is appreciated.

user26857
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james
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    Welcome to MSE. A question should be written in such a way that it can be understood even by someone who did not read the title. – José Carlos Santos Dec 04 '21 at 10:41
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    Appreciate the feedback I will rewrite it... – james Dec 04 '21 at 11:03
  • It says, $f(x)$ is monotonically increasing function iff $f'(x)$ is positive over the domain of $f(x)$ thus if $$f(x) = \ln(1+e^x) \implies f'(x) = \frac {e^x}{1+e^x}$$ here, $f'$ is positive over the domain now it's easy to find that $\ln(1+e^x)$ can't have negative value as $\ln(u)$ is positive if $u>1$ thus as $f(x\to -\infty) = \to 0$ – Darshan P. Dec 04 '21 at 11:19
  • "has a minimum at $-\infty$" is nonsense. $h$ has no minimum value on $\mathbb{R}$. – K.defaoite Dec 04 '21 at 11:30
  • I agree with @K.defaoite As, OP said "monotonically increasing function" means $h(x-\delta) < h(x)$ so for every $x$ there must exist $x-\delta$ – Darshan P. Dec 04 '21 at 11:42

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