Evaluating maximum Could someone help me understand how to solve this question? First off, you must know that I ask questions on Stack Exchange only after doing my best to solve it first and I'm not trying to find easy answers for my homework. Also, since I'm really fond of math, I hate not being able to answer a question.
Q: Find the range of $(x+1)(x+2)(x+3)(x+4)$ in the interval $[-6,6]$.
My try: I saw that since $f(x)$ is strictly increasing in the interval $[0,6]$ the maximum is clearly 5040. Also, the minimum has to be something negative, since the function has four zeroes in the interval $[-6,0]$. However, I'm unable to calculate the minimum.
Could someone help me in evaluating the range? I'd greatly appreciate an answer that would show me a general method applicable even for fourth- or higher-degree expressions(I can handle quadratics and cubics pretty easily.) 
I don't want to know just the answer for this question, I'd like to know the general method, since I want to develop my knowledge, be self-reliant and not limit myself to specific questions. 
P.S.: I have this idea that one must solve questions using only the concepts known at that level (for example, if I can help it, I never solve a congruency problem using similarity of triangles) so try to include an answer that uses calculus or algebra. That said, I'd like innovative answers too.
 A: Hint:
Set 
$(x+1)(x+4)=u$
$(x+2)(x+3)=u+2$
A: Perform the following substitution:
$u=x-5/2$.  On doing this your function will become symmetrical about the y-axis. Your equation becomes 
$$(u-3/2)(u-1/2)(u+1/2)(u+3/2)$$
over the range $[-7/2,17/2]$. The natural question is what did we gain out of this? By symmetry above I meant that the function is now an even function i.e $$f(-u) = f(u)$$
Therefore, you only have to bother with the values on $[0,17/2]$, as the behavior on $[-7/2,0]$ is same as that on $[0,7/2]$. The other advantage is that taking derivative is easy as you will now see.
$$f(u)= (u^2-1/4)(u^2-9/4) $$
hence
$$f^{'}(u) = 2u(2u^2-5/2) $$
At minimum value(as we already know the maximum because of your insights :p) $$f^{'}(u) = 0 \implies u=0,\sqrt5/2,-\sqrt5/2$$
We only need to compare $f(0)$ and $f(\sqrt5/2)$(as $f$ is an even function).Without evaluating you can see that $f(0) > 0$, thus the minimum value is at $f(\sqrt5/2) = (5/4 -1/4)(5/4 - 9/4) = -1$
Hence the range of $f$ on $[-6,6]$ is $[-1,5040]$
