Finding consecutive integers to satisfy the IVT to find roots. Is there a way to do this algebraically, or is it all guess work? I was hoping someone could verify my sanity here. I vaguely recall from college a way to solve this algebraically instead of just picking and choosing numbers to satisfy the conditions of the Intermediate Value Theorem (IVT).
"Find consecutive integers $a$ and $b$ so that $f(x) = x^4 - 2x^3 - 3x + 2$ is guaranteed to have a root between $a$ and $b$ by the IVT."
What we did was just choose $f(0)$, $f(1)$, $f(2)$, and $f(3)$. I ended up finding a root between $[0, 1]$ and $[2, 3]$. But I can not for the life of me remember if there's a way to do this without just randomly choosing numbers and narrowing it down. 
I wasn't positive if there was a way to do this algebraically, it's been a while.
 A: All of the purely precalculus methods will involve a bit of "guess and test" (using the Rational Roots Theorem, Playing around with Linear Factors, etc.) Your best bet would be a little bit of Calculus. Here are a couple methods:
Method 1 -- Newton-Raphson Method. 
If you don't know, this is a way to approximate the root of a function using the derivative. $x_n-\frac{f(x_n)}{f'(x_n)}$
This will, however, involve a little bit of that undesirable guessing you were talking about. 
First, we find the derivative of $f(x)=x^4-2x^3-3x+2$.
$f'(x)=4x^3-6x^2-3$
Then, we take a guess at a root. Let's use $3$. (This is an iterative method by the way -- it will approach the root as the number of iterations gets larger)
$3-\frac{f(3)}{f'(3)}= 2.607$
$2.607-\frac{f(2.607)}{f'(2.607)}= 2.425$
If you kept going with the iterations, you would eventually approach the root -- 2.381. And here comes more guessing and testing -- finding the values to use for the IVT. Pretty much all methods will involve some light guesswork. For this method, just look until you find a value lower than the root and one higher than the root.
Method 2 -- Playing with Derivatives
Remember that $f'(x)=4x^3-6x^2-3$. Let's try and use this to our advantage. $f(x)$ has a critical point at $1.746$ (we found this by solving $f'(x)=0$). Now, remember that a root of odd multiplicity "crosses" the axis. The root we found just happens to be of odd multiplicity, therefore it will cross the x-axis. With this new knowledge, let's test the critical point that we found using the first derivative test (is it a maximum or minimum). $f'(1)=-5 $ and $f'(2)= 2$. The derivative goes from negative to positive at the critical point, so the critical point is a minimum. What is the minimum? Quick answer: the lowest point on the function. If the function crosses the x-axis, it, therefore, has a point $c$ such that the sign of $f(c)$ is negative. Now, you can just evaluate this function at higher points such that the sign becomes positive and use the IVT to prove the existence of a root.
In conclusion, finding the values for the IVT usually involves a bit of guessing and testing. These two methods show semi-algebraic ways to find the points but introduce unneeded extra work. 
