Question

Find the values of $x$, for which the function $f\left(x\right)=x^3+12x^2+36x+6$ is increasing.

Answer 1

Let

$f\left(x\right)=x^3+12x^2+36x+6$

Differentiate it w.r.t $x$

$f^{\prime }\left(x\right)=3x^2+24x+36$

The function increases when $f^{\prime }\left(x\right)>0$

So,

$3x^2+24x+36>0$

$x^2+8x+12>0$

$x^2+6x+2x+12>0$

$x(x+6)+2(x+6)>0$

$(x+2)(x+6)>0$

$x<-2$

$x<-6$

So, for $x<-2$ and $x<-6$ the function is increasing\