<s>[INST] If the range of a function is $[a,b]$ ,where $a,b \in \mathbb N$ ,find $f(a+b)$. Let  $f(x)= (x+1) (x+2) (x+3) (x+4) + 5$  where $x \in [-6,6]$. If the range of this function is $[a,b]$ ,where $a,b \in \mathbb  N$ ,find $f(a+b)$.
ATTEMPT:
Sketching the graph the we know that minima of this function is between $-1$ and $-2$, and since the function is strictly increasing in the interval $[-1,6]$, maxima will be at $6$.
But how to get the minima?
 [/INST] $\bf{My\; Solution::}$ Let $\displaystyle \left(x+\frac{1}{2}\right) = t\;,$ and $\displaystyle -\frac{11}{2}\leq t \leq \frac{13}{2}$. 
Then expression convert into 
$$\displaystyle f(t) = \left(t-\frac{3}{2}\right)\cdot \left(t-\frac{1}{2}\right)\cdot \left(t+\frac{1}{2}\right)\cdot \left(t+\frac{3}{2}\right)=\left(t^2-\frac{9}{4}\right)\cdot \left(t^2-\frac{1}{4}\right)+5$$
So $$\displaystyle f(t) = t^4-\frac{5}{2}t^2+\frac{9}{16}+5$$ and $$\displaystyle0 \leq t^2\leq \frac{169}{4}$$
So we get $$\displaystyle f(t) = \left(t^2-\frac{5}{4}\right)^2+\frac{89}{16}-\frac{25}{16}\Rightarrow f(t) = \left(t^2-\frac{5}{4}\right)^2+4$$ and $$\displaystyle0 \leq t^2\leq \frac{169}{4}$$
So $\displaystyle \bf{f(t)_{Min.} = 4}$ when $\displaystyle t^2=\frac{5}{4}$ 
And $\displaystyle \bf{f(t)_{max.} = \left(\frac{169}{4}-\frac{5}{4}\right)^2+4=(41)^2+4 = 1681+4 = 1685,}$ When $\displaystyle t^2= \frac{169}{4}$
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