Finding the number of roots of a function in a given interval Consider the following question:

Show that the function has exactly one root in the interval $[-2,-1]$.
  $$f(x)=x^4+3x+1.$$

I can solve it algebraically, but I'm supposed to use Rolle's and the Mean value theorem.
The value of $f(-2)$ is positive and $f(-1)$ is negative. So that proves that at least one zero exists. I do not understand how I can prove the exact amount of zeroes in the interval is $1$. For all I know, the function might cross the $x$-axis number of times before reaching $-1$.
 A: Consider the derivative
$$
f'(x)=4x^3+3 = -1 + 4(x^3+1)
$$
over the given interval (esp. its sign).

The polynomial is close to $$g(x)=(x^3+3)(x+\tfrac13)=x^4+\tfrac13x^3+3x+1,$$ so that its roots will be found close to $x=-\tfrac13$, $x=-\sqrt[3]3$ and $x=\sqrt[3]3·\frac12(1\pm i\sqrt3)$.
\begin{array}{l|l}
\text{roots of }g(x) & \text{roots of }f(x) \\ \hline
   -0.33333333                 &  -0.33766677  \\
   -1.44224957                 &  -1.30748610   \\
   0.721124785 + 1.249024766 i &   0.82257643 + 1.26031796 i  \\
   0.721124785 - 1.249024766 i &   0.82257643 - 1.26031796 i  \\
\end{array}
So we see that a perturbation of about $10\%$ (of a maximum of $3$) in the coefficients leads to a perturbation of about $10\%$ in the roots. Here this is small enough to leave the distribution of real and complex roots and the qualitative location (quadrants) invariant.
A: First,
The problem has provided you with an interval: $[-2, -1]$
As you described above, you would first plug in these end points and determine if there is a root.
Therefore, solving:
$$f(-2) = (-2)^4 + 3*(-2) + 1 = 11$$
and
$$f(-1) = (-1)^4 + 3*(-1) + 1 = -1$$
As you can see, since one endpoint is positive and the other is negative, by the Intermediate Value Theorem, there is at least one zero in this interval.
Now, proving there is only $1$ zero, that requires the derivative. Using the derivative allows to see whether the function is increasing or decreasing on that interval, which allows us to prove whether or not there is only $1$ zero in the interval.
Taking the derivative, we get:
$$
f'(x)=4x^3+3
$$
Now, first note any critical points, or where $f'(x) = 0$. This allows us to see where $f(x)$ has a maximum.
Here we can see that $x^3 = -\frac{3}{4}$, or $x = \left(-\frac{3}{4}\right)^{1/3}$ is the only critical point. Next we look at the interval and plug in the endpoints to see the behavior of $f(x)$.
When $-2 \le x \le 1$, $f'(x) < 0$, therefore $f(x)$ is monotone decreasing on this interval.
Therefore, since $f(x)$ is monotone decreasing on this interval, you can prove that $f(x)$ has only $1$ zero in that interval.

Alternatively, you can prove that through the Mean Value Theorem on this interval that $x = \left(-\frac{3}{4}\right)^{1/3}$  is a critical point, and also that there exists no critical point in the interval. which leads you through the same steps above:
A: As noted in the answer of LutzL, in this case the derivative is simple and shows that there is only one stationary point in $x=\sqrt[3]{-\frac{3}{4}}$ that is not in the interval $[-2,-1]$, so in this interval there is only one zero of the function.
But, in general, the roots of the two polynomials $f(x)$ and $f'(x)$ can be difficult to find, and in such a case we can find the number of roots in a given interval, using the Sturm's theorem.
A: $f(x)=x^4+3x+1$ has $0$ change-sign and $f(-x)=x^4-3x+1$ has $2$ ones therefore, because of $4-2=2>0$, $f$ has at least $2$ imaginary roots, hence at most $2$ real roots. The derivative is 
$f’(x)=4x^3+3$ so the function has only one real extrema at the point $x_0=-\sqrt[3]\frac 34=-0,908..$ (which $-1<x_0$)
and since $x^4$ the function has exactly two real roots (and two imaginary roots). On the other hand $f(-2)>0$ and $f(-1)<0$ and $f(0)=1>0$ so there is a real root in $[-2,-1]$ the other root is in the interval [-1,0]
NOTE.-The criteria used to determine there are at least $2$ imaginary roots has been atributed to Descartes (it is not the known Rule of Descartes but a variation of it). I asked in StackExchange to prove this criteria but my post was put in hold for some one.
