Increasing cubic function $f(x)$ in $3 < x < 4$, $f(2) = 8$ and $f(-4) = -16$ How do we solve the following problem?

A function $f(x)$ is a cubic function which is increasing at interval $3 < x < 4$. If $f(2) = 8$ and $f(-4) = -16$, find the equation of function!*

I try to solve it as 3 variable equation by assuming that $f'(x)$ is $0$ at $3$ and $4$. But I don't know how to put the fact that the function is increasing at $3 < x < 4$ to something that I can solve using algbera. Can somebody help explain this problem to me? 
 A: you have a lot of freedom
$f(x) = ax^3 + bx^2 + cx + d$
If we set $b = 0$ and $a,c > 0$
Then $f(x)$ will be increasing for all $x$
We still have 3 unknowns and 2 conditions to solve for, meaning we still have freedom.  We can assign a value to $a$ that is convenient and solve for $c, d$
let $a = \frac 14$
$f(x) = \frac 14 x^3 + c(x+4)$
Now $f(x)$ will go through the point $(-4,-16)$
And we just need to solve for $c$
$f(x) = \frac 14 x^3 + x+4$
And this is just one of many equations you could choose.
Update:
$f(x) = ax^3 + bx^2 + cx + d\\
f'(x) = 3ax^2 + 2bx+c > 0$ 
When $x\in [3,4]$
Either $a > 0$ and $f'(x)$ has no real roots.
or $a>0$ and both roots are to the left of $3$, or both roots are to the right of $4$
or $a<0$ and [3,4] is inside the roots.
$x = \frac {b \pm \sqrt {b^2 - 3ac}}{3}$
My short-cut was to assume $a>0$ and $b = 0$ thus $f'(x)$ has no real roots and $f(x)$ is monotonically increasing.
If you want to solve for one of the other cases, you are welcome to, but you are asking for more work... 
