# Determine the monotonic intervals of this function y=2x^3-6x^2-18x-7.

Determine the monotonic intervals of the functions
i need to determine the monotonic intervals of this function $y=2{x}^{3}-6{x}^{2}-18x-7$. I tried the below but i am not sure if i am doing it right.
My work: $\begin{array}{rl}y=2{x}^{3}-6{x}^{2}-18x-7& ⟺6{x}^{2}-12x-18=0\\ & ⟺6\left({x}^{2}-2x-3\right)=0\\ & ⟺\left(x-3\right)\left(x+1\right)\\ & ⟺x-3=0x+1=0\\ & ⟺x=3,x=-1\end{array}$
so my function increases when $x\in \left[3,+\mathrm{\infty }\left[$ and decreases when $x\in \left[-1,3\right]\cup \right]-\mathrm{\infty },-1\right]$.
Please i want to know how to solve this problem any help with explanation will be appreciated. thanks in advanced
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Favatc6
Step 1
You have correctly found the derivative $\frac{dy}{dx}=6{x}^{2}-12x-18=6\left({x}^{2}-2x-3\right)=6\left(x-3\right)\left(x+1\right)$ and where it is zero, but you have not quite got the intervals correct.
The derivatives is positive if and only if $\left(x-3\right)\left(x+1\right)>0$ which is positive if and only if i.e. if and only if .
Step 3
So the function is (strictly) increasing on the interval $\left(-\mathrm{\infty },-1\right]$ and on the interval $\left[3,\mathrm{\infty }\right)$.
The function is (strictly) decreasing on the interval $\left[-1,3\right]$