Consider the function $ m(x) = 12x^5 + 60x^{12} $. Differentiate $ m $ and use the derivative to determine each of the following:

1. All intervals on which $ m $ is increasing. If there is more than one interval, separate them with a comma. Use open intervals and exact values.
$ m $ increases on:

2. All intervals on which $ m $ is decreasing. If there is more than one interval, separate them with a comma. Use open intervals and exact values.
$ m $ decreases on:

3. The value(s) of $ x $ at which $ m $ has a relative maximum. If there is more than one solution, separate them with a comma. Use exact values.
$ m $ has relative maximum(s) at:

4. The value(s) of $ x $ at which $ m $ has a relative minimum. If there is more than one solution, separate them with a comma. Use exact values.
$ m $ has relative minimum(s) at:

To determine the intervals on which the function m(x) = 12x^5 + 60x^2 is increasing or decreasing, we need to find the derivative of m(x) and analyze its sign.

First, let's find the derivative of m(x) using the power rule for differentiation:

m'(x) = 60x^4 + 120x

To determine the intervals on which m(x) is increasing, we need to find where m'(x) > 0.

Setting m'(x) > 0:

60x^4 + 120x > 0

We can factor out 60x:

60x(x^3 + 2) > 0

Now, we have two factors to consider:

60x > 0: This is true for x > 0.

x^3 + 2 > 0: To determine the sign of this factor, we can analyze the sign of x^3 + 2 for various intervals.

When x < -∛2, x^3 + 2 < 0.

When -∛2 < x < 0, x^3 + 2 > 0.

When x > 0, x^3 + 2 > 0.

Combining these intervals, we find that x^3 + 2 > 0 for x ≠ -∛2.

Therefore, m(x) is increasing on the intervals (-∞, -∛2) and (0, ∞).

To determine the intervals on which m(x) is decreasing, we need to find where m'(x) < 0.

Setting m'(x) < 0:

60x^4 + 120x < 0

Again, we can factor out 60x:

60x(x^3 + 2) < 0

Analyzing the sign of the two factors:

60x < 0: This is true for x < 0.

x^3 + 2 < 0: We can use the same intervals as before to determine the sign of this factor.

When x < -∛2, x^3 + 2 < 0.

When -∛2 < x < 0, x^3 + 2 > 0.

When x > 0, x^3 + 2 > 0.

Combining these intervals, we find that x^3 + 2 < 0 for -∛2 < x < 0.

Therefore, m(x) is decreasing on the interval (-∛2, 0).

To find the relative maximum and minimum points, we need to find where the derivative equals zero or is undefined.

Setting m'(x) = 0:

60x^4 + 120x = 0

Factoring out 60x:

60x(x^3 + 2) = 0

This gives us two solutions: x = 0 and x = -∛2.

Therefore, m(x) has a relative maximum at x = 0 and no relative minimum.

In summary:

m(x) is increasing on the intervals (-∞, -∛2) and (0, ∞).

m(x) is decreasing on the interval (-∛2, 0).

m(x) has a relative maximum at x = 0 and no relative minimum.

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