Answer :
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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