We will have the following:
First, we are given:
[tex]f(x)=-2x^3+36x^2-162x+2[/tex]Now, we determine the critical points:
[tex]\begin{gathered} f^{\prime}(x)=-6x^2+72x-162=0 \\ \\ \Rightarrow x=\frac{-(72)\pm\sqrt{(72)^2-4(-6)(-162)}}{2(-6)}\Rightarrow \\ x=9 \\ x=3 \end{gathered}[/tex]So, we can see that the expression has two critical points at x = 3 & x = 9.
So, at the points:
[tex](3,-214)[/tex]&
[tex](9,2)[/tex]Now, we determine which one is the inflection point as follows:
[tex]f^{\prime}^{\prime}(x)=-12x+72[/tex]So, we analyze the function at the points given:
*x = 3:
[tex]f^{\prime\prime}(3)=-12(3)+72\Rightarrow f^{\prime\prime}(3)=36[/tex]So, at x there is a local minimum.
*x = 9:
[tex]f^{\prime}^{\prime}(9)=-12(9)+72\Rightarrow f^{\prime}^{\prime}(8)=-36[/tex]So, at x = 9 there is a local maximum.
The expression given has no inflection point.
