JEE Main · 2021 · 31 Aug 2021 · Shift 2 · Application of Derivatives

Write $f(x)=a{x}^3+b{x}^2+cx+d$, so $f^{\prime }(x)=3a{x}^2+2bx+c$ and $f^{\prime \prime }(x)=6ax+2b$. A local minimum of the quadratic $f^{\prime }$ sits at its vertex, where $f^{\prime \prime }=0$; this occurs at $x=-1$, giving $-6a+2b=0$, so $b=3a$. A local minimum of $f$ at $x=1$ gives $f^{\prime }(1)=3a+2b+c=0$, hence $c=-9a$. The data $f(1)=-5a+d=-10$ and $f(-1)=11a+d=6$ subtract to $16a=16$, so $a=1$, $b=3$, $c=-9$, $d=-5$. Thus $f(x)=x^3+3x^2-9x-5$ and $f(3)=27+27-27-5=22$.