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Consider the function [tex]f(x) = -36x^5 + 60x^3 - 10[/tex].

Differentiate [tex]f[/tex] and use the derivative to determine each of the following:

1. All intervals on which [tex]f[/tex] is increasing. If there is more than one interval, separate them with a comma. Use open intervals and exact values.

- [tex]f[/tex] increases on: ________

2. All intervals on which [tex]f[/tex] is decreasing. If there is more than one interval, separate them with a comma. Use open intervals and exact values.

- [tex]f[/tex] decreases on: ________

3. The value(s) of [tex]x[/tex] at which [tex]f[/tex] has a relative maximum. If there is more than one solution, separate them with a comma. Use exact values.

- [tex]f[/tex] has relative maximum(s) at [tex]x = [/tex] ________

4. The value(s) of [tex]x[/tex] at which [tex]f[/tex] has a relative minimum. If there is more than one solution, separate them with a comma. Use exact values.

- [tex]f[/tex] has relative minimum(s) at [tex]x = [/tex] ________

Answer :

The intervals on which f is increasing are (-∞, -1) and (0, 1). The interval on which f is decreasing is (-1, 0). The function has a relative maximum at x = -1 and a relative minimum at x = 1.

To determine the intervals on which the function f(x) = [tex]-36x^5 + 60x^3 -[/tex]10 is increasing or decreasing, we need to find the derivative of the function and analyze its sign.

First, let's find the derivative of f(x):

f'(x) =[tex]-180x^4 + 180x^2[/tex]

To determine the intervals where f is increasing, we look for values of x where f'(x) > 0.

[tex]-180x^4 + 180x^2 > 0[/tex]

Factoring out common terms, we have:

[tex]180x^2(-x^2 + 1) > 0[/tex]

The quadratic expression [tex]-x^2 + 1[/tex]can be factored as -(x - 1)(x + 1). Thus, we have:

[tex]180x^2(x - 1)(x + 1) > 0[/tex]

To determine the sign of the expression, we use a sign chart:

| (-∞) | -1 | 0 | 1 | (∞) |

180x^2 | + | + | 0 | + | + |

(x - 1) | - | - | - | + | + |

(x + 1) | - | - | - | - | + |

Product | - | + | 0 | - | + |

From the sign chart, we can see that f'(x) > 0 when x < -1 or 0 < x < 1.

Therefore, f is increasing on the intervals (-∞, -1) and (0, 1).

To determine the intervals where f is decreasing, we look for values of x where f'(x) < 0.

From the sign chart, we can see that f'(x) < 0 when -1 < x < 0.

Therefore, f is decreasing on the interval (-1, 0).

To find the relative maximum and minimum, we need to find the critical points. Critical points occur when f'(x) = 0 or when f'(x) is undefined.

Setting f'(x) = 0:

[tex]-180x^4 + 180x^2 = 0[/tex]

Factoring out common terms, we have:

[tex]180x^2(x^2 - 1) = 0[/tex]

This equation is satisfied when x = 0 or x = ±1.

Therefore, the relative maximum occurs at x = -1 and the relative minimum occurs at x = 1.

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