$0<f(1)<f(3)<f(2)$ Limits, Continuity, Maxima and minima

Question

Given the function $f(x)$ that fulfills the following $0<f(1)<f(3)<f(2)$ determine if the statements are true or false.

1. if the function is defined in the interval $[1,3]$ then we will conclude that $\lim \limits_{x \to2}f(x)>0$
2. if the function is continuous in $[1,3]$ we can conclude that there is a point $c$ in the interval $(1,3)$ that is $f'(c)=0$
3. if the function is differentiable in $[1,3]$ we can conclude that $f'(2)=0$
4. if the function is defined in $[1,3]$ then there is no Vertical asymptotes
5. if the function is differentiable in $[1,3]$ we can conclude that $x=2$ is a local maxima or global maxima

What I tried:

1. I think that it is true according to the information given $0<f(1)<f(3)<f(2)$ and $\lim \limits_{x \to2}f(x)$ is $f(2)$ and they are all greater than zero.
2. In part 2 I thought about two things, the first thing was that the statement is not true because even if it is continuous it does not mean that the function is necessarily differentiable so that makes it false. But the second thing I thought about was the Extreme value theorem but even with that it still does not mean that $f′(c)=0$ so my guess is that the statement is false(?)
3. I also believe that this is false, this rang a bell for Rolle's theorem at first we need it to be differentiable in $(1,3)$ and continuous in $[1,3]$ and we need $f(a)=f(b)$ and we don't have that here according to $0<f(1)<f(3)<f(2)$ but I am not sure.
4. This seems true to me because they said it is defined in $[1,3]$
5. Also seems true according to Extreme value theorem and the information given $0<f(1)<f(3)<f(2)$.

I usually have more information on what I tried on my posts but honestly, on this question I found it very hard to even begin with, I would very much appreciate tips and hints on how to approach these questions.

Answer 1

1. One can construct a function $f$ such that $f(x_0)$ is defined but $\lim_{x\to x_0} f(x)$ is not defined. Consider $sign(x)$ at point $x_0=0$. There are also functions for which $\lim_{x\to x_0} f(x)$ defined but is not equal to $f(x_0)$. Consider $f$ where $f(0) = 0$ and $f(x) = 1$ for every $x \neq 0$ at point $x_0=0$.

2. You are correct. Continuity of $f$ does not imply differentiability. Though if we assume $f$ to be differentiable, the statement is true. Apply the intermediate value theorem to interval $[1,2]$: let $a \in [1,2]$ be such that $f(a)=f(3)$. Then apply Rolle's theorem to interval $[a, 3]$.

3. One can easily provide a contradicting example. Just draw three points $(1, f(1))$, $(2, f(2))$, $(3, f(3))$ on a plane such that $0<f(1)<f(3)<f(2)$ and draw a curve through them that doesn't have a horizontal tangent at point $x=2$. For a rigorous construction, a parabola $y=ax^2+bx+c$ with $a<0$ and its vertex $x_0 \in (2,3)$ might be a good choice -- the details are up to you.

4. Think of a function $f$ that is defined at $[1,2)\cup (2,3]$ and $\lim_{x\to 2} f(x)=\pm \infty$. You then obviously have a vertical asymptote at point $x=2$. Setting $f(2)=0$ does not do anything to asymptote.

5. An example contradicting 3. will also probably contradict 5.