Suppose $f$ is continuous on $\mathbb R$ and $f(0)=1$ and $\lim_{x\to\infty} f(x)=0$. Show that $f$ must have a max/min on $\mathbb R$

Question

I am a bit stuck here. How can I show $f$ have a max or min? Any hints are greatly appreciated. Thanks.

My thoughts are as follows: since $\lim_{x\to\infty} f(x)=0$ when it goes to infinity. There must be horizontal asymptote and $f(0)=1$ is min? I am not sure how to proceed. Thanks for your help.

Answer 1

This is false. $e^{-x}$ satisfies the requirements but does not have a maximum, because $\lim_{x\rightarrow -\infty} e^{-x}=\infty$.

If you also ask that $\lim_{x\rightarrow -\infty} f(x)=0$, then, there exists M>0 large enough so that $|f(x)|<1$ for all $x\in \mathbb{R}\setminus [-M,M]$. As $[-M,M]$ is compact and $f$ is continuous, the Extreme Value Theorem implies that $f$ attains its maximum in $[-M,M]$, as $f(0)=1$, we know that this maximum is equal or greater than $1$, so it is also the maximum of $f$ in $\mathbb{R}$.

The function doesn't have to have a minimum. e.g. $f(x)=e^{-x^2}$.