If $y=f^{\prime}(x)$ is a cubic polynomial having maxima at $x=-1$ and $f^{\prime}(x)$ has minima at

Question

If $y=f^{\prime}(x)$ is a cubic polynomial having maxima at $x=-1$ and $f^{\prime}(x)$ has minima at $x=1$, also $f(-1)=10$, $f(1)=-6$. The square of the distance between the points, which are maxima or minima is

Professor PC Thomas Classes · Answer

Let $f(x)=a x^{3}+b x^{2}+c x+d$ as $f(-1)=10$ and $f(1)=-6$ (given)\ therefore $-a+b-c+d=10$\ and $a+b+c+d=-6$\ From (i) and (ii)\ $\Rightarrow \quad b+d=2$\ and $a+c=-8$\ Again $f^{\prime}(x)=3 a x^{2}+2 b x+c$\ At $x=-1, f(x)$ is maximum\ $\Rightarrow \quad 3 a-2 b+c=0$\ $	herefore \quad f^{\prime \prime}(1)=0$\ $\Rightarrow \quad 6 a+2 b=0$\ Using (iii), (iv), (v) and (vi), we get\ $a=1, b=-3, c=-9, d=5$\ $	herefore \quad f(x)=x^{3}-3 x^{2}-9 x+5$\ $f^{\prime}(x)=3 x^{2}-6 x-9=3(x+1)(x-3)$\ Obviously $x=3$ is a point of minima.\ Coordinates of point of maxima and minima are $(-1,10)$ and $(3,-22)$ respectively.\ $	herefore \quad$ Distance between them\ $=\sqrt{(3+1)^{2}+(-22-10)^{2}}$\ $=\sqrt{1040}=4 \sqrt{65}$\