Respuesta :

Given:

The function is:

[tex]y=2x^3-15x^2+36x-20[/tex]

To find:

The turning points.

Solution:

We have,

[tex]y=2x^3-15x^2+36x-20[/tex]

Differentiate the given function with respect to x.

[tex]y'=2(3x^2)-15(2x)+36(1)-(0)[/tex]

[tex]y'=6x^2-30x+36[/tex]

[tex]y'=6(x^2-5x+6)[/tex]

For turning point, [tex]y'=0[/tex].

[tex]6(x^2-5x+6)=0[/tex]

[tex]x^2-3x-2x+6=0[/tex]

[tex]x(x-3)-2(x-3)=0[/tex]

[tex](x-3)(x-2)=0[/tex]

Using zero product property, we get

[tex]x-3=0[/tex] and [tex]x-2=0[/tex]

[tex]x=3[/tex] and [tex]x=2[/tex]

Therefore, the turning points of the given function are at [tex]x=2[/tex] and [tex]x=3[/tex].