Question

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 $(a+b)$.

Answer 1

Answer: 5049

$f(x)=\left(x^2+5x+4\right)\left(x^2+5x+6\right)+5$
$=\left[\left(x^2+5x+5\right)-1\right]\left[\left(x^2+5x+5\right)+1\right]+5$
$={\left(x^2+5x+5\right)}^2-1+5$
$f(x)={\left(x^2+5x+5\right)}^2+4$
Hence $f(x)$ has a minimum value 4 when $x^2+5x+5=0$
i.e. $x=\frac{-5\pm \sqrt{5}}{2};$ Hence $x=\frac{-(5+\sqrt{5})}{2}\in [-6,6]$ also maximum occurs at $x=6$
${f(x)|}_{\max }=(36+30+5{)}^2+4=(71{)}^2+4=5041+4=5045$
range is $[4,5045]$
$\therefore a=4;b=5045\Rightarrow a+b=5049$
Alternatively: $f(x)-5=g(x)]$