Prove that $f(x)$ is a constant if it has two derivatives and the second derivative is greater than or equal to 0.
I know that if $f(x) = const$, its 1st derivative should equal 0, but how can it be equal to 0 and increasing??
EDIT: Forgot to add that $f(x)$ is also less than or equal to 0, 😀
Best Answer
If the second derivative is non-negative, the first derivative must be weakly increasing. It is clear that if $f'(x) \rightarrow a>0$ as $x \rightarrow \infty$ imply that $f$ is eventually positive, so we can assume that $f'(x)<0$ everywhere.
Now, $f'(x)\rightarrow b\leq 0$ as $x \rightarrow -\infty$, since $f'$ is increasing. (The case $b=-\infty$ can also be a possibility).
If $b=0$, then the function is constant, and if $b<0$, then a similar argument as above shows that $f(x) \rightarrow \infty$ as $x \rightarrow -\infty$. This contradicts negativity of $f$, so $f$ must be a negative constant.