JEE Main · 2020 · 08 Jan 2020 · 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 critical point of $f$ at $x=-1$ gives $3a-2b+c=0$, and a critical point of $f^{\prime }$ at $x=1$ gives $f^{\prime \prime }(1)=6a+2b=0$, so $b=-3a$ and then $c=-9a$. The values $f(-1)=10$ and $f(1)=-6$ give $5a+d=10$ and $-11a+d=-6$; subtracting, $16a=16$, so $a=1$ with $b=-3$, $c=-9$, $d=5$. Then $f^{\prime }(x)=3x^2-6x-9=3(x+1)(x-3)$, with critical points at $x=-1$ and $x=3$, and since $f^{\prime \prime }(3)=6(3)-6=12>0$, the function has a local minimum at $x=3$.