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i need to determine the monotonic intervals of this function $y=2x^3-6x^2-18x-7$. I tried the below but i am not sure if i am doing it right.

My work: begin{align} y=2x^3-6x^2-18x-7
&Longleftrightarrow 6x^2-12x-18=0\
&Longleftrightarrow6(x^2-2x-3)=0\
&Longleftrightarrow(x-3) (x+1)\
&Longleftrightarrow x-3=0 x+1=0\
&Longleftrightarrow x=3, x=-1\
end{align}

so my function increases when
$xin[3, +infty[$ and decreases when $xin[-1, 3]cup ]-infty, -1]$

Please i want to know how to solve this problem any help with explanation will be appreciated. thanks in advanced

You have correctly found the derivative

$$frac{dy}{dx}=6x^2-12x-18=6(x^2-2x-3)=6(x-3)(x+1)$$

and where it is zero, but you have not quite got the intervals correct.

The derivatives is positive if and only if

$$(x-3)(x+1)>0$$

which is positive if and only if

$$(x>3 text{ and } x>-1)quad text{ or }quad (x<3 text{ and } x<-1)$$

i.e. if and only if

$$(x>3)quad text{ or }quad (x<-1)$$

So the function is (strictly) increasing on the interval $(-infty, -1]$ and on the interval $[3,infty)$.

The function is (strictly) decreasing on the interval $[-1,3]$.