To solve this, you have to know that the first derivative of a function is its slope. When an interval is increasing, it has a positive slope. Thus, we are trying to solve for when the first derivative of a function is positive/negative.
f(x)=2x^3+6x^2-18x+2
f'(x)=6x^2+12x-18
f'(x)=6(x^2+2x-3)
f'(x)=6(x+3)(x-1)
So the zeroes of f'(x) are at x=1, x=-3
Because there is no multiplicity, when the function passes a zero, he y value is changing signs.
Since f'(0)=-18, intervals -3<x<1 is decreasing(because -3<0<1)
Thus, every other portion of the graph is increasing.
Therefore, you get:
Increasing: (negative infinite, -3), (1, infinite)
Decreasing:(-3,1)
I think you multiply length times width if not I am sorry. I think the answer is 30. Sorry if it isn’t
Answer:
1/6
Step-by-step explanation:
The gradient of the line segment is the same as slope
m = (y2-y1)/(x2-x1)
= ( -4 - -5)/( 2 - -4)
= (-4+5)/( 2 +4)
1/6
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