The first derivative of the function f is defined by f′(x) = sin(x3 − x) for 0 ≤ x ≤ 2 . On what intervals is f increasing?

We need to remember that the derivative of a function can be used in order to determine where a function is increasing or decreasing on a specific domain.

If f′(x) > 0 at each point in an interval a, the function is increasing on a.

If f′(x) < 0 at each point in an interval a, the function is decreasing on a.

Solution to the problem

On this case we have the following derivate:

[tex] f'(x) = sin (x^3 -x) , 0 \leq x \leq 2[/tex]

We can find the critical points:

[tex] sin (x^3 -x) = 0[/tex]

[tex] x(x^2 -1) = 0[/tex] or [tex] x(x^2 -1) = \pi[/tex]

So then the critical points are x=0, x=1. and x=1.691 and now we can evaluate the function on any point of the following intervals (0,1),(1,1.691), (1.691,2)

If we find the second derivate for the function we got:

[tex] f''(x) = (3x^2 -1) cos (x^3 -x)=0[/tex]

[tex] cos(x^3 -x) = 0[/tex]

[tex] x^3 -x = n \frac{\pi}{2}[/tex]

[tex] x(x^2-1)=n \frac{\pi}{2} [/tex]

We have a maximum at (1.44,1) and a minimum at (0.58,-0.38). so the interval of increase is 0.58 to 1

Between (0,0.578) we have this:

[tex] f'(0.5) = -0.366 <0 [/tex] So then the function is decreasing on the interval (0,0.578)

Between (0.578,1) we have this:

[tex] f'(0.7) = -0.349 >0 [/tex] We have that f'(0.7)>f'(0.5)

So then our function is incrasing at (0.578,1)

If we select a value between 1 and 1.445 we got:

[tex] f'(1.3) = 0.781 >0 [/tex] So then the function is increasing on the interval (1,1.445)

We have a maximum at (1.445,1) and a minimum at (0.58,-0.38). so the interval of increase is 0.58 to 1.445

And the answer for this case would be [tex]0.58\leq x \leq 1.445[/tex]

joaobezerra joaobezerra · Answer 2 · 2021-11-25T15:13:59+01:00

Using the first derivative rule, it is found that the function f is increasing on the interval (1, 1.69).

The first derivative rule states that:

When the derivative f'(x) is positive, the function is increasing.
When the derivative f''(x) is negative, the function is decreasing.

In this problem, the derivative of the function is given by:

[tex]f^{\prime}(x) = \sin{(x^3 - x)}, 0 \leq x \leq 2[/tex]

The graph of this derivative is sketched at the end of the answer.

Using the interval, it is positive on the interval (1, 1.69), hence, this is the interval in which the function f is increasing.

A similar problem is given at https://brainly.com/question/13539822