Samantha V.

asked • 03/13/20

Find the local extrema of the function f (x) = x^3 + 3x^2 + 1

Claytonia B.

To find local extrema, you would need to find the first derivative of the function. Using the power property, we get 3x^2 + 6x. If you set that equal to zero (because first derivative represents slope, and the slope of the tangent line at a local extrema is zero), you can factor it and solve. In factored form its 3x(x+2)=0 . Setting each factor to zero and solving gets x=0 and x=-2. Substitute each x value back into the original to get a complete ordered pair. When you substitute zero for x, (0)^3+3(0)^2+1 = 1. So that ordered pair is (0,1). When you substitute -2, (-2)^3+3(-2)^2+1 = 5 so the ordered pair is (-2,5)
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03/15/20

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Raymond B. answered • 03/13/20

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take the first derivative, set it equal to zero and solve for x


f'(x) = 3x2 + 6x = 0


factor and set each factor = 0


3x(x+2) = 0


3x= 0 then x =0 is a local minimum


x+2 = 0 so x=-2 is a local maximum


plug those into the original cubic function to get the y coordinates of the local extrema


f(0) = 1 so (0,1) is the local minimum


f(-2) = -8 +12 +1 = 5 so (-2,5) is the local maximum



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