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What are the increasing intervals of the graph -2x^3-3x^2+432x+1

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Answer:

decreasing: (-∞, -9) ∪ (8, ∞)

increasing: (-9, 8)

Explanation:

You want the intervals where the function f(x) = -2x³ -3x² +432x +1 is increasing and decreasing.

Derivative

The slope of the graph is given by its derivative:

f'(x) = -6x² -6x +432 = -6(x +1/2)² +433.5

Critical points

The slope is zero where ...

-6(x +1/2)² = -433.5

(x +1/2)² = 72.25

x +1/2 = ±8 1/2

x = -9, +8

Intervals

The graph will be decreasing for x < -9 and x > 8, since the leading coefficient is negative. It will be increasing between those values:

decreasing: (-∞, -9) ∪ (8, ∞)

increasing: (-9, 8)

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Additional comment

A cubic (or any odd-degree) function with a positive leading coefficient generally increases over its domain, with a possible flat spot or interval of decrease. When the leading coefficient is negative, the function is mostly decreasing, with a possible interval of increase, as here.

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What are the increasing intervals of the graph -2x^3-3x^2+432x+1-example-1
User Edward Sun
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