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The last answer - 9x squared y squared
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Answer:
OH MY GOD I NEEED HELP WITH QUESTION TOOOOOOO
First, you need to find the derivative of this function. This is done by multiplying the exponent of the variable by the coefficient, and then reducing the exponent by 1.
f'(x)=3x^2-3
Now, set this function equal to 0 to find x-values of the relative max and min.
0=3x^2-3
0=3(x^2-1)
0=3(x+1)(x-1)
x=-1, 1
To determine which is the max and which is the min, plug in values to f'(x) that are greater than and less than each. We will use -2, 0, 2.
f'(-2)=3(-2)^2-3=3(4)-3=12-3=9
f'(0)=3(0)^2-3=3(0)-3=0-3=-3
f'(2)=3(2)^2=3(4)-3=12-3=9
We examine the sign changes to determine whether it is a max or a min. If the sign goes from + to -, then it is a maximum. If it goes from - to +, it is a minimum. Therefore, x=-1 is a relative maximum and x=1 is a relative miminum.
To determine the values of the relative max and min, plug in the x-values to f(x).
f(-1)=(-1)^3-3(-1)+1=-1+3+1=3
f(1)=(1)^3-3(1)+1=1-3+1=-1
Hope this helps!!
