Question

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?

Answer 1

Answer:

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 $0.58\leq x \leq 1.445$

Step-by-step explanation:

Previous concepts

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:

$f'(x) = sin (x^3 -x) , 0 \leq x \leq 2$

We can find the critical points:

$sin (x^3 -x) = 0$

$x(x^2 -1) = 0$ or $x(x^2 -1) = \pi$

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:

$f''(x) = (3x^2 -1) cos (x^3 -x)=0$

$cos(x^3 -x) = 0$

$x^3 -x = n \frac{\pi}{2}$

$x(x^2-1)=n \frac{\pi}{2}$

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:

$f'(0.5) = -0.366$ So then the function is decreasing on the interval (0,0.578)

Between (0.578,1) we have this:

$f'(0.7) = -0.349 >0$ 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:

$f'(1.3) = 0.781 >0$ 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 $0.58\leq x \leq 1.445$