# What will x=3 be in this equation

Assume f is differentiable and has just one critical point, at x = 3. In parts (a) - (d), you are given additional conditions. In each case decide whether x = 3 is a local maximum, a local minimum, or neither. You should be able to sketch possible graphs for all four cases.

we have this equation ) $$\lim_{x\rightarrow \infty}{f(x)}=\infty\\ \lim_{x\rightarrow -\infty}{f(x)} = \infty$$

will that make x=3 a local minimum or maximum ? I think it will be a local minimum like saying $$-x^2$$ amd I right ?

• Are you joking? You only have the limit behavior. That says nothing about $x=3$. Jun 26 '14 at 6:50
• @RainiervanEs fixed Jun 26 '14 at 6:53

With the conditions you have added, an absolute (and therefore local) minimum is forced at $x=3$. We give a proof, though it does not appear as if you are expected to. It may be you are just being asked to use your geometric intuition, which in this case led to the right answer.

Let $f(3)=p$. Since $\lim_{x\to-\infty}f(x)=\infty$, there is a negative number $a$ such that $f(a)\gt p$. Since $\lim_{x\to\infty}f(x)=\infty$, there is a $b\gt 3$ such that $f(b)\gt p$.

Since our function is continuous, it attains an absolute minimum (and an absolute maximum) on the closed interval $[a,b]$. The minimum value(s) cannot be taken at an endpoint, so it (they) must be attained at a point or points $q$ in the interval $(a,b)$. If a maximum or minimum is attained at such an internal point $c$, and the derivative exists everywhere, we must have $f'(c)=0$.

However, we are told that $3$ is the only point $c$ such that $f'(c)=0$. So we achieve a local (and absolute) minimum at $3$ and nowhere else.

From those two limits, you cannot infer that $x = 3$ is a local minimum/maximum. The function $x\mapsto x^2$ satisfies the equation, but $x = 3$ is nothing.

It's indeed a local minimum.

We can take $y=(x-3)^2$ as an example, since it has the same limits and the same critical point at $(3,0)$.