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Answer :
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].
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