if i have two limits going to negative infinity and a function that is differentiable , does that mean there has to exist a maximum?

Question

Suppose that $f(x)$ is differentiable for every $x\in\Bbb R$ and that $$\lim_{x\to \pm\infty}f(x) = -\infty$$, then there exist a maximum point $\in\Bbb R$

so i was thinking if i can use Wiestrass theorem...

Answer 1

As $\lim_{x\to \pm\infty}f(x) = -\infty$, you can find $M>0$ such that $f(x) < f(0)$ for $x \notin [-M;M]$. Then as $f$ in continuous on $[-M,M]$ it reaches a maximum at say $y \in [-M,M]$ which is also a global maximum because $f(x) < f(0) \le f(y)$ for $y \notin [-M,M]$.

Answer 2

First notice that $$\lim_{x\to \pm\infty}f(x) = -\infty$$ is equivalent to saying $\forall M<0, \exists x_0>0 $ such that $f(x)<M \ \forall x\geq x_0$ or $x\le -x_0$. by definition.

In the light of this, we can "crop" down the domain of $\mathbb R$ into $I=[-x_0,x_0]$ to look for the maximum, knowing that if there is one, it must be in $I$.

Then, since differentiability on a interval implies continuity, we conclude that $f(x)$ is continuous on the closed interval $I$.

Naturally, a closed, bounded interval with a continuous function $f(x)$ defined on it must have $f$ bounded on $I$. (Boundedness theorem)

Finally, since $f$ is bounded on closed interval $I$, we conclude there is an absolute maximum value on $I$. (Extremum value theorem) Thus, f reaches a maximum value.

QED.