Question

Let $f\left(x\right)=(x+1)(x+2)(x+3)(x+4)+5$ 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

Given $f\left(x\right)=(x+1)(x+2)(x+3)(x+4)+5$ for $x\in [-6,6]$

Take $g\left(x\right)=(x+1)(x+2)(x+3)(x+4)$

Rearranged as $g\left(x\right)=(x+1)(x+4)(x+2)(x+3)$

$=(x^2+5x+4)(x^2+5x+6)$

$=(x^2+5x+4)(x^2+5x+4+2)$

$=t(t+2)$ where $t=x^2+5x+4$

$=t^2+2t+1-1$

$={(t+1)}^2-1$

$={(x^2+5x+4+1)}^2-1$ where $t=x^2+5x+4$

$={(x^2+5x+5)}^2-1$

Let $g\left(x\right)=x^2+5x+5$

$\Rightarrow f\left(x\right)={(x^2+5x+5)}^2-1+5={(x^2+5x+5)}^2+4$

Range of $x^2+5x+5$ for $x\in [-6,6]$

Discriminant$=5^2-4\times 1\times 5=25-20=5>0$

$\therefore$ there exists two roots

Let $y=x^2+5x+5$

$\Rightarrow y^{\prime }=2x+5$

$\Rightarrow y^{\prime }=0\Rightarrow 2x+5=0$

$\Rightarrow 2x=-5$

$\therefore x=\frac{-5}{2}=-2.5\in [-6,6]$

Substitute $x=\frac{-5}{2}$ in $x^2+5x+5$

we get $\frac{25}{4}-\frac{25}{2}+5=\frac{25-50+20}{4}=\frac{-5}{4}$

$\therefore$ minimum value $=\frac{-5}{4}$

Put $x=-6$ in $x^2+5x+5$

$\Rightarrow 36-30+5=11$

Put $x=6$ in $x^2+5x+5$

$\Rightarrow 36+30+5=71$

$\therefore$ Range of $x^2+5x+5$ is $[\frac{-5}{4},71]$

Range of ${(x^2+5x+5)}^2\in [0,{71}^2]$ since $t^2\ge 0$

$\therefore$ Range of ${(x^2+5x+5)}^2\in [0,5041]$

Range of ${(x^2+5x+5)}^2+4\in [0,5041]+4=[0,5045]$

$\therefore$ range of $f\left(x\right)=[0,5045]$

where $a=4,b=5045$

$\therefore a+b=4+5045=5049$