Considering the critical points of the function, we have that:
- The function is increasing for |x| > 1.63.
- The function is decreasing for |x| < 1.63.
What are the critical points of a function?
The critical points of a function are the values of x for which:
[tex]f^{\prime}(x) = 0[/tex]
In this problem, the function is:
[tex]f(x) = 3x^3 - 16x + 2[/tex]
The derivative is:
[tex]f^{\prime}{x} = 6x^2 - 16[/tex]
The critical points are given as follows:
[tex]6x^2 - 16 = 0[/tex]
[tex]x^2 = \frac{16}{6}[/tex]
[tex]x = \pm \sqrt{\frac{16}{6}}[/tex]
[tex]x = \pm 1.63[/tex]
For x < -1.63, one example of the derivative is:
[tex]f^{\prime}{-2} = 6(-2)^2 - 16 = 8[/tex]
Positive, hence increasing.
For -1.63 < x < 1.63, one example of the derivative is:
[tex]f^{\prime}{0} = 6(0)^2 - 16 = -16[/tex]
Negative, hence decreasing.
For x > 1.63, one example of the derivative is:
[tex]f^{\prime}{2} = 6(2)^2 - 16 = 8[/tex]
Positive, hence increasing.
Hence:
- The function is increasing for |x| > 1.63.
- The function is decreasing for |x| < 1.63.
More can be learned about critical points at https://brainly.com/question/2256078
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