On the domain , we know that the derivative begins positive, and because the concavity is positive, we know the derivative is increasing. Thus, the derivative stays positive for this entire interval, and the function increases from 2 to 4. Thus,  must be greater than .

In the case of the interval , we know that the derivative is increasing, but it starts out negative. Thus, perhaps the derivative only increased from -1 to -0.5 in this interval, and the function would have decreased the entire time. In this case,  would be less than , so we can't really say anything about these values.

For the remaining two, there's not any clear way to relate the functions at  and . While we know  needs to be bigger than 1, we don't know by how much. Similarly, while we know  needs to be bigger than -1, we don't know by how much. Thus, it's completely possible that  and .

As for  and , we know even less. In between the two intervals, our function could have shot up a million, or shot down by the same amount. Thus, there's no safe comparison we can make between these two values.

To find the intervals where the function is increasing, we need to find the points at which its slope changes from positive to negative and vice versa. The first derivative, which is the slope at any point, will help us. 

First, we find the derivative of , using the power rule for each term. Recall that the power rule says

Also, the constant multiple rule will apply to the coefficients of each term. The constant multiple rule simply says that any constant factor of a term will "carry" to the derivative of that term. For example:

Lastly, the derivative of a constant is zero. This will result in the last term, , dropping off as we take the derivative.

Applying these rules, we find the derivative

Where the derivative is positive (blue line in graph), the tangent line to the original function is angled up. Where the derivative is negative (red line in graph), the slope of the tangent line is angled down.

The points where the tangent line's slope transitions from negative to positive or vice versa, are called the critical points. At these points, the tangent line becomes a horizontal line with a slope of zero (green line in graph). In other words, the "critical points" occur when the derivative is zero. These points will be the endpoints of our intervals of increasing and decreasing. To find the critical points, we will set the derivative equal to zero and solve for x. In this problem factoring is the best method:

Now that we have found the critical points, we need to know whether the original function is increasing or decreasing in the intervals between them. We will do so by testing a point in each interval and determining whether the derivative is positive or negative at that point. This is called the first derivative line test.

For the interval , we will test .

(Note: I will use the factored form of the derivative, but we could also use the polynomial version. Both will give the same result)

Since the derivative is negative at this point, we can conclude that the derivative is negative for the whole interval. Thus, the original function is decreasing on .

For the interval , we will test .

Since the derivative is positive at this point, we can again conclude that the derivative is positive for the whole interval. Thus, the original function is increasing on .

For the interval , we will test .

Since the derivative is negative at this point, we can again conclude that the derivative is negative for the whole interval. Thus, the original function is decreasing on .

Lastly, for the interval , we will test .

Since the derivative is positive at this point, we know that the derivative is positive for the whole interval. Thus the original function is increasing on .

From these 4 results, we now know the answer. The function is increasing on the intervals .

The rate of change of y is also the derivative of y.

Differentiate the function given. 

You should get 

If p(t) gives the position of an asteroid as a function of time, find the function which models the velocity of the asteroid as a function of time.

Begin by recalling that velocity is the first derivative of position. So all we need to do is find the first derivative of our position function.

Recall that the derivative of sine is cosine, and that the derivative of polynomials can be found by multiplying each term by its exponent, and decreasing the exponent by 1.

Starting with:

We get:

Given j(k), find the rate of change when k=5

Let's begin by realizing that a rate of change refers to a derivative.

So, we need to find the derivative of j(k)

We find this by multiplying each term by the exponent, and decreasing the exponent by 1

Next, plug in 5 to find our answer:

So, our rate of change is -221.

If p(t) gives the position of a planet as a function of time, find the function which models the planet's velocity.

Velocity is the first derivative of position.

Therefore, all we need to do to solve this problem is to find the first derivative.

We can do this via the power rule and the rule for differentiating sine.

1) 

2) 

So, we these rules in mind, we get:

So our final answer is:

If p(t) gives the position of a planet as a function of time, find the planet's velocity when t=0.

Velocity is the first derivative of position. Therefore, all we need to do to solve this problem is to find the first derivative of p(t) and then plug in 0 for t and solve.

We can do this via the power rule and the rule for differentiating sine and cosine.

1) 

2) 

So, we these rules in mind, we get:

So our velocity function is:

Now, plug in 0 and simplify.

So, our answer is -11. We are not given any units, so we do not need to worry about them.

The rate of change of a function is given by the derivative of that function. So, to find the rate of change of production, we must take the first derivative of the function for production, which is equal to

found using the following rules:

, 

The rate of change of the concrete flow is given by the first derivative of the concrete flow function:

and was found using the following rules:

, , 

To determine the speed of the car, we must take the first derivative of the position function, which gives us the rate of change of the position of the car - in other words, speed. 

The derivative was found using the following rule:

, 

Evaluating the derivative function at t=5, we get our speed as