Master Intermediate Value Theorem Problems

Use the Intermediate Value Theorem to prove that the equation
$-<!--\; -\; -->x^3+4\sin <!--\; \; -->(x)+4{\cos }^2<!--\; \; -->(x)=0$
has at least two solutions. You should carefully justify each of the hypothesis of the theorem.

How do I know that at least two exist because I know you're meant to look for a change in sign.

On a recent problem, I received the following scenario: An object moves back and forth on a straight track. During the time interval $0\le <!--\; \le \; -->t\le <!--\; \le \; -->30$ minutes, the object's position, $x$, and velocity, $v$, are continuous functions; some of their values are shown in the table (which I have reproduced below).

$\begin{array}{|ccc|}\hline t\mathbf{\text{ (min)}}& x(t)\mathbf{\text{ (feet)}}& v(t)\mathbf{\text{ (feet/min)}}\\ 0& \text{12}& -<!--\; -\; -->20\\ 10& \text{50}& \text{20}\\ 15& \text{18}& \text{3}\\ 25& \text{60}& -<!--\; -\; -->2\\ 30& \text{60}& \text{10}\\ \hline\end{array}$

The question was, for $0<t<30$, does there exist a time t when $v(t)=-<!--\; -\; -->22$?

I tried to apply the intermediate value theorem, and concluded that the answer was "not necessarily." I reasoned that because the values of $v(t)$ in the table ranged between −20 (at the beginning) and 10 (at the end), and −22 wasn't between those two values, we couldn't be sure that such a t exists.

The teacher gave me 3/4 points on the question. Their comment was that I should have considered the mean value theorem, too. They did not write anything about my intermediate value theorem analysis, but I still have grave doubts about my application of the intermediate value theorem ... and, as for the mean value theorem, I have no idea how to proceed.

Would anyone here be able to shed some light on how the intermediate value and mean value theorems could be used to determine whether there exists a t where $v(t)=-<!--\; -\; -->22$? Thanks so much.

Suppose that $f:[a,b]\to <!--\; \to \; -->\mathbb{R}$ is continuous and that $f(a)<0$ and $f(b)>0$. By the intermediate-value theorem, the set $S=\{x\in <!--\; \in \; -->[a,b]:f(x)=0\}$ is nonempty. If $c=supS$, prove that $c\in <!--\; \in \; -->S$.

My first thought was to show that $S$ is finite therefore $c\in <!--\; \in \; -->S$, but there is no guarantee that $f$ doesn't have infinitely many zeros.

A thought I have now is to show that $c>max(S)$ can not be true, but I do not know how to show this, is this even the correct thing to show? Thank you in advance for any input.