9514 1404 393
Answer:
Step-by-step explanation:
I find it convenient to draw the graph first when looking for relative extrema.
The function can be differentiated to get ...
f'(x) = -3x^2 +9
This is zero when ...
-3x^2 +9 = 0
x^2 = 3
x = ±√3 . . . . . x-values of relative extrema
Then the extreme values are ...
f(±√3) = x(9 -x^2) = (±√3)(9 -3) = ±6√3
The lower extreme (minimum) corresponds to the lower value of x (-√3), so the extrema are ...
(x, y) = (-√3, -6√3) and (√3, 6√3)
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Since the leading coefficient is negative and the degree is odd, the function is decreasing for values of x below the minimum and above the maximum. It is increasing for values of x between the minimum and the maximum.
decreasing: x < -√3, and √3 < x
increasing: -√3 < x < √3
Answer:
The correct option is 1. The area of cross section area is 48 mm².
Step-by-step explanation:
From the find it is noticed that the cross section is a rectangle with length 4 mm and width is 12 mm.
The area of a rectangle is the product of its dimensions.
Where, l is length of the rectangle and w is width of the rectangle.
The area of cross section is
Therefore the area of cross section area is 48 mm². Option 1 is correct.
