Question

Consider the function below. (If an answer does not exist, enter DNE.) F(x) = x 9 − x (a) Find the interval of increase. (Enter your answer using interval notation.)

Answer 1

`(-\infty, -0.76) \cup (0.76, \infty)`

Explanation:

The first step to solve this question is finding the critical points of the function F(x), which are x for which:

`F'(x) = 0`

In this question:

`F(x) = x^(9) - x`

So

`F'(x) = 9x^(8) - 1`

`9x^(8) - 1 = 0`

`9x^(8) = 1`

`x^(8) = (1)/(9)`

`x = \sqrt[8]{(1)/(9)}`

`x = \pm 0.76`

So we have three intervals:

`(-\infty, -0.76), (-0.76, 0.76), (0.76, \infty)`

We take a value of x from each interval. If the derivative is positive, the function increases. Otherwise, it decreases.

First interval:

`(-\infty, -0.76)`

Will take x = -1.

`F'(-1) = 9*(-1)^(8) - 1 = 9 - 1 = 8`

Positive, so increases.

Second interval:

(-0.76, 0.76),

Will take x = 0;

`F'(0) = 9*(0)^(8) - 1 = 0 - 1 = -1`

Negative, so decreases

Third interval:

`(0.76, \infty)`

Will take x = 1

`F'(1) = 9*(1)^(8) - 1 = 9 - 1 = 8`

Positive, so increases.

Interval of increase:

First and third, so:

`(-\infty, -0.76) \cup (0.76, \infty)`