Question

If $P\left(x\right)$ be a polynomial of degree $3$ satisfying $p(-1)=10,p\left(1\right)=-6$ and $p\left(x\right)$ has maxima at $x=-1$ and $p\left(x\right)$ has minima at $x=1$. Find the distance between the local maxima and local minima of the curve.

Answer 1

Let $p\left(x\right)=a{x}^3+b{x}^2+cx+d$
$p(-1)=10\Rightarrow -a+b-c+d=10$
$p\left(1\right)=-6\Rightarrow a+b+c+d=-6$
$p\left(x\right)$ has max. at $x=-1\Rightarrow p^{\prime }(-1)=0$
$\Rightarrow 3a-2b+c=0$
Solving $\left(1\right),\left(2\right),\left(3\right)$ and $\left(4\right)$, we get
From $\left(4\right),b=-3a$
from $\left(3\right),3a+6a+c=0\Rightarrow c=-9a$
from $\left(2\right)a-3a-9a+d=-6\Rightarrow d=11a-6$
from $\left(1\right),-a-3a+9a+11a-6=10$
$\Rightarrow 16a=16\Rightarrow a=1$
$\Rightarrow b=-3,c=-9,d=5$
$\Rightarrow wp\left(x\right)=x^3-3x^2-9x+5$
$p^{\prime }\left(x\right)=0\Rightarrow 3x^2-6x-9=0$
$\Rightarrow 3(x+1)(x-3)=0$
$\Rightarrow x=-1$ is a point of maxima (given)
and $x=3$ is a point of minima
[$\because$ maxima and minima occur alternatively]
$\therefore$ point of local maxima is $(-1,10)$ and local minima is $(3,-22)$
And the distance between them is
$=\sqrt{[3-(-1){]}^2+(-22-10{)}^2}$
$=\sqrt{16+1024}$
$=\sqrt{1040}=4\sqrt{65}$