Let $f(x)=(x+1)(x+2)(x+3)(x+4)+5$ where $x\in[-6,6]$. If the range of the function is [a,b] where $a,b\in N$, then find the value of (a+b) Let $f(x)=(x+1)(x+2)(x+3)(x+4)+5$ where $x\in[-6,6]$. If the range of the function is [a,b] where $a,b\in N$, then find the value of (a+b)
My attempt is as follows:-
Rewrite $f(x)=g(x)+5$ where $g(x)=(x+1)(x+2)(x+3)(x+4)$
Let's find the maximum value of g(x)
It can be clearly seen that from $x=-6$ to $x=6$, maximum value of $g(x)$ is at $x=6$.
$$g(6)=5040$$
Let's find the minimum value of $g(x)$
From the sign scheme one can see that negative value of $g(x)$ occurs in the interval $(-4,-3)$ and $(-2,-1)$
Hence intuitively it feels that the minimum value of g(x) would be at $x=-\dfrac{3}{2}$ or at $x=-\dfrac{7}{2}$ as in case of parabola also, minimum value is at average of both the roots.
So $g\left(-\dfrac{3}{2}\right)=-\dfrac{1}{2}\cdot\dfrac{1}{2}\cdot\dfrac{3}{2}\cdot\dfrac{5}{2}=-\dfrac{15}{8}$
At $x=-\dfrac{7}{2}$, $g\left(-\dfrac{7}{2}\right)=-\dfrac{5}{2}\cdot\dfrac{3}{2}\cdot\dfrac{1}{2}\cdot\dfrac{1}{2}=-\dfrac{15}{8}$
So range of $g(x)$ would be $[\dfrac{-15}{8},5040]$
Hence range of $f(x)$=$\left[\dfrac{25}{8},5045\right]$
But it is given that the range is $[a,b]$ where $a,b\in N$
I am stuck here.  I am also not able to prove mathematically that at $x=-\dfrac{3}{2}$ or $x=-\dfrac{7}{2}$, minimum value of $g(x)$ will occur.
Please help me in this.
 A: Like you have suggested, consider the function $$g(x)=(x+1)(x+2)(x+3)(x+4).$$
$g'(x)$ potentially has minimum values at its critical numbers, which we can find by setting the derivative equal to $0$.
\begin{align}
g(x)&=(x+1)(x+2)(x+3)(x+4)\\
\log g(x)&=\log\big((x+1)(x+2)(x+3)(x+4)\big)\\
\log g(x)&=\log(x+1)+\log(x+2)+\log(x+3)+\log(x+4)\\
\frac{d}{dx}\big(\log g(x)\big)&=\frac{d}{dx}\big(\log(x+1)+\log(x+2)+\log(x+3)+\log(x+4)\big)\\
\frac{g'(x)}{g(x)}&=\frac1{x+1}+\frac1{x+2}+\frac1{x+3}+\frac1{x+4}\\
g'(x)&=g(x)\bigg(\frac1{x+1}+\frac1{x+2}+\frac1{x+3}+\frac1{x+4}\bigg)\\
g'(x)&=(x+1)(x+2)(x+3)(x+4)\bigg(\frac1{x+1}+\frac1{x+2}+\frac1{x+3}+\frac1{x+4}\bigg)
\end{align}
Solving the equation $$\frac1{x+1}+\frac1{x+2}+\frac1{x+3}+\frac1{x+4}=0$$ gives us $x=-\dfrac52-\dfrac{\sqrt5}2$, $-\dfrac52$, and $-\dfrac52+\dfrac{\sqrt5}2$.  To find maximum and minimum values on an interval, we then need to compare the values at these critical numbers and at the bounds of the interval.
The maximum value is $g(6)=5040$ (as you found) and the minimum value is $g(-\frac52+\frac{\sqrt5}2)=-1$.
Since $f(x)=g(x)+5$, the range of $f(x)$ on $[-6,6]$ is therefore $[-1+5,5040+5]$ or $\boxed{[4,5045]}.$  As a result, the value of $a+b$ is $\boxed{5049}$.
A: The function is symmetric about the line $x = -2.5$
We could write
$f(x) = ((x+2.5) + 1.5)((x+2.5)+0.5)((x+2.5)-0.5)((x+2.5) - 1.5) + 5$
Which equals
$f(x) = ((x+2.5)^2 - 1.5^2)((x+2.5)^2-0.5^2) + 5$
Now lets do the substituion $u = (x+2.5)^2$
$(u - 2.25)(u-0.25) + 5$
Which has its minimum when $u = \frac {5}{4}$
$(\frac {5}{4} - 2.25)(\frac {5}{4}-0.25) + 5 = 4$
A: Continuing discussion in comments, use first derivative test.
$$
f’(x)=2x^3+15x^2+35x+25=(2x+5)(x^2+5x+5)
$$
Now the roots for the quadratic are
$x = \frac{-5+\sqrt{5}}{2}, \frac{-5-\sqrt{5}}{2}$. So to check for maximum and minimum value, check $f(-6)=120$, $f(\frac{-5+\sqrt{5}}{2})=-1$, $f(\frac{5}{2})>0$ cannot be max/min by your discussion, $f(\frac{-5-\sqrt{5}}{2})=-1$ and $f(6)=5040$. So we just take the max/min now.
A: $$f(x)=(x^2+5x+4)(x^2+5x+6)+5=(x^2+5x+5)^2-1+5\geq4.$$
The equality occurs for $x^2+5x+5=0,$ which happens on $[-6,6],$ which says that $4$ is a minimal value.
Now, $f(6)=5045$ and since for any $-6\leq x\leq6$
$$5045-f(x)=(6-x)(x+11)(x^2+5x+76)\geq0,$$ we see that $5045$ is a maximal value.
A: The range will be f(X)$\in$[4,5045]
It is simple approach will be solution to quadratic equation as in article. https://www.mathsdiscussion.com/solution-of-quadratic-equation/
