JEE Main · 2021 · 24 Feb 2021 · Shift 2 · Application of Derivatives

For $x<-5$ the derivative is $f^{\prime }(x)=-55<0$, so $f$ decreases on that ray. On $-5\le x\le 4$, $f^{\prime }(x)=6x^2-6x-120=6(x-5)(x+4)$, which is positive only for $x<-4$ or $x>5$, and inside this piece that leaves $-5<x<-4$. For $x>4$, $f^{\prime }(x)=6x^2-6x-36=6(x-3)(x+2)$, and both factors are positive once $x>4$, so $f^{\prime }>0$ throughout. Collecting the stretches on which the derivative is positive, $f$ is increasing exactly on $(-5,-4)\cup (4,\infty )$.