Use the Mean Value Theorem to estimate $f(8)−f(3)$ Suppose $f(x)$ is continuous on $[3,8]$ and $−4≤f′(x)≤3$ for all $x$ in $(3,8)$.
The mean value theorem to me means that there is a given $f(x)$, but here there is no function given. I'm not sure how to work out $f(a)$ and $f(b)$ when there's no $f(x)$ to plug them into. 
 A: MVT claims that if $f$ is a continuous function over $[a,b]$, then $f'(c)=\frac{f(b)-f(a)}{b-a}$ for some $c \in (a,b)$. Since you have $-4 \leq f'(x) \leq 3 \quad \forall x \in (3,8)$, then you know that $-4 \leq f'(c) \leq 3$ for $c$ mentioned in the MVT.
Intuitively, MVT claims that the "average speed (slope, change)" over the entire domain equal to some "instant speed (slope, change)" at some point $c$ inside the domain. Another intuitive way to understand this is that draw a line $AB$ through point $A=(a,f(a))$ and point $B=(b, f(b))$, you can draw a tangent line $\ell$ of the function $f$ at some point $c$ such that $\ell \parallel AB$.
A: since is given that $-4\le f'(x)\le3\forall x\in(3,8)$ and $f(x)$ is continuous then by mean value theorem we have that exist some $\xi\in(3,8)$ such that
$$f'(\xi)=\frac{f(8)-f(3)}{8-3}\\
f'(\xi)=\frac{f(8)-f(3)}{5}$$
then since $\xi\in(3,8)$ you can use the above inequality to conclude that
$$-4\le\frac{f(8)-f(3)}{5}\le3\\
-20\le f(8)-f(3)\le15$$
