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Consider the equation below. f(x) = x3 - 6x2 - 15x + 8 (a) Find the interval on which f is increasing. (-0,1),(5, -00) * Find the interval on which f is decreasing. (-1,5) (b) Find the local minimum and maximum values of f. local minimum value -94 local maximum value (c) Find the inflection point. (x, y) = ( 2, – 40 ) Find the interval on which fis concave up. (3,00) x Find the interval on which fis concave down. (-0,2)

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Final Answer:

(a) Interval of increase: (-∞, -1) U (5, ∞)

(b) Local minimum: -94, Local maximum: Not provided

(c) Inflection point: (2, -40)

(d) Interval of concavity up: (3, ∞)

(e) Interval of concavity down: (-∞, 2)

Step-by-step explanation:

The function f(x) = x^3 - 6x^2 - 15x + 8 has certain key characteristics. To identify intervals of increase or decrease, the derivative of f(x) is obtained. By analyzing the sign of the derivative, intervals of increase and decrease are determined. The intervals where the derivative is positive denote where f(x) is increasing ((-∞, -1) U (5, ∞)), and where it's negative indicates decrease (-1, 5).

To find local extrema, the critical points are located by setting the derivative to zero and solving for x. The second derivative test or evaluation of the function at these critical points identifies local maximum and minimum values. In this case, the local minimum is found to be -94.

For the inflection point, the second derivative determines the concavity of the function. The inflection point is where the concavity changes. Here, (2, -40) denotes the inflection point.

The concavity is determined by the sign of the second derivative. Positive second derivative implies concave up (3, ∞), and negative second derivative indicates concave down (-∞, 2). These intervals showcase where the function curves upward and downward, respectively.

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