Answer:
Option A.
Step-by-step explanation:
The given equation is
[tex]y=-x^3-5x^2-3x+9[/tex] ...(1)
We need to find the local maximum.
Differentiate the given equation w.r.t. x.
[tex]y=-(3x^{3-1})-5(2x^{2-1})-3(1)+0[/tex]
[tex]y'=-3x^2-10x-3[/tex] ...(2)
Now, equate y'=0 to find the extrem points.
[tex]-3x^2-10x-3=0[/tex]
[tex]-3x^2-9x-x-3=0[/tex]
[tex]-3x(x+3)-1(x+3)=0[/tex]
[tex]-(3x+1)(x+3)[/tex]
[tex]3x+1=0\Rightarrow x=-\dfrac{1}{3}[/tex]
[tex]x+3=0\Rightarrow x=-3[/tex]
Differentiate (2) w.r.t. x.
[tex]y''=-6x-10[/tex]
For [tex]x=-\dfrac{1}{3}[/tex],
[tex]y''=-6(-\dfrac{1}{3})-10=2-10=-8<0[/tex] maximum
For [tex]x=-3[/tex],
[tex]y''=-6(-3)-10=18-10=8>0[/tex] minimum
So, the given equation has local maximum at x=-1/3 and the maximum value is
[tex]y=-(-\dfrac{1}{3})^3-5(-\dfrac{1}{3})^2-3(-\dfrac{1}{3})+9\approx 9.48[/tex]
The local maximum at (-0.33,9.48).
Hence, the correct option is A.
