Can someone please help me with the answer.

Mathematics

1 answer:

Elza [17]4 years ago

5 0

Answer:

The answer to your question is : f(-2) = 2

Step-by-step explanation:

Just look for -2 in the x- axis and look for some line or point, vertical to number two.

In your example you can find a point two units upward the x-axis.

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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?

defon

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:

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