To determine the intervals at which the function is increasing or decreasing, we first need to calculate its derivative:
[tex]\begin{gathered} f^{\prime}(x)=3\cdot3x^2+2\cdot9x-27 \\ f^{\prime}(x)=9x^2+18x-27 \end{gathered}[/tex]
We need to find the values of the roots of the expression above:
[tex]\begin{gathered} x=\frac{-18\pm\sqrt[]{18^2-4\cdot9\cdot(-27)}}{2\cdot9} \\ x=\frac{-18\pm\sqrt[]{1296}}{18} \\ x=\frac{-18\pm36}{18} \\ x_1=\frac{-18+36}{18}=\frac{18}{18}=1 \\ x_2=\frac{-18-36}{18}=\frac{-54}{18}=-3 \end{gathered}[/tex]
We can rewrite the expression for the derivative as:
[tex]f^{\prime}(x)=9\cdot(x-1)\cdot(x+3)[/tex]
Therefore the derivative is positive when the signs of "x-1" and "x+3" are equal, therefore there are two possibilities, that they are both negative and both positive. We have:
[tex]\begin{gathered} x-1>0 \\ x>1 \\ x-1<0 \\ x<1 \\ x+3>0 \\ x>-3 \\ x+3<0 \\ x<-3 \end{gathered}[/tex]
They are both negative when x<-3 and both positive when x>1.
So the function is increasing at the intervals:
[tex]\begin{gathered} (-\alpha,-3) \\ (1,\alpha) \end{gathered}[/tex]
And decreasing on the interval:
[tex](-3,1)[/tex]
