f(x) = -x(x-3)^2

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We have the function as

f(x)=-x(x-3)^2

Determinig the x intercept ,put f(x)=0

-x(x-3)^2=0

x=0 and x=3

y intercept is put x=0 we get 0

Thus the determined point are (3,0)(0,0)

Also -x(x-3)^2 >0

x(- (-infinity,0)

And -x(x-3)^2 <0

x(- (0,infinity)

Determining relative the maxima and minima value

d(f(X)/dx=0

d(-x^3-9x+6x^2)/dx=0

-3x^2-9+12x=0

Divide with -3 ,we get

x^2-4x+3=0

x^2-3x-x+3=0

x(x-3)-1(x-3)=0

(x-1)(x-3)=0

x=1 and x=3

Now the second derivative is d(f(X)/dx^2=d(-3x^2-9+12x)/dx=-6x+12

at x=1 d^y/dx^2=6>0 therefore minima occur and the minimum value is -4

at x=3,d^2y/dx^2=-6*3+12 =-6 maxima occur and the maximum value is 3

Using the above information creating a table of values

x y

0 0

3 0

1 -4

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