Respuesta :
Answer:
[tex](-\infty, -0.76) \cup (0.76, \infty)[/tex]
Step-by-step explanation:
The first step to solve this question is finding the critical points of the function F(x), which are x for which:
[tex]F'(x) = 0[/tex]
In this question:
[tex]F(x) = x^{9} - x[/tex]
So
[tex]F'(x) = 9x^{8} - 1[/tex]
[tex]9x^{8} - 1 = 0[/tex]
[tex]9x^{8} = 1[/tex]
[tex]x^{8} = \frac{1}{9}[/tex]
[tex]x = \sqrt[8]{\frac{1}{9}}[/tex]
[tex]x = \pm 0.76[/tex]
So we have three intervals:
[tex](-\infty, -0.76), (-0.76, 0.76), (0.76, \infty)[/tex]
We take a value of x from each interval. If the derivative is positive, the function increases. Otherwise, it decreases.
First interval:
[tex](-\infty, -0.76)[/tex]
Will take x = -1.
[tex]F'(-1) = 9*(-1)^{8} - 1 = 9 - 1 = 8[/tex]
Positive, so increases.
Second interval:
(-0.76, 0.76),
Will take x = 0;
[tex]F'(0) = 9*(0)^{8} - 1 = 0 - 1 = -1[/tex]
Negative, so decreases
Third interval:
[tex](0.76, \infty)[/tex]
Will take x = 1
[tex]F'(1) = 9*(1)^{8} - 1 = 9 - 1 = 8[/tex]
Positive, so increases.
Interval of increase:
First and third, so:
[tex](-\infty, -0.76) \cup (0.76, \infty)[/tex]
