Is the function b(t) concave up, concave down, or neither when t is equal to -3?

To test for concavity, we need to find the sign of the function's second derivative at the given time.

Begin by recalling that the derivative of a polynomial is found by multiplying each term by its exponent, then decreasing the exponent by 1.

Doing this gets us the following:

Almost there, but we need b"(-3)

b"(-3) is negative, therefore our function is concave down when t=-3

To determine the intervals on which the function is concave down, we must find the intervals on which the second derivative is negative.

First, we must find the second derivative of the function:

The derivatives were found using the following rules:

, , 

Note that for the first rule, the chain rule, as it applies to the natural log, regardless of the constant in front of x, the derivative of the natural log will always be the same.

Next, we must find the values at which the second derivative is equal to zero. This never occurs, so there is no place at which the second derivative is equal to zero. Furthermore, the second derivative is always positive, so the function is never concave down.

To find concavity, you need to find the second derivative. In order to find that, you must first determine the first derivative. When taking the derivative, multiply the exponent by the coefficent in front of the x and then subtract one from the exponent. The first derivative is . Then, find the second derivative is . Set that equal to 0 to get your critical point: Then, test values on either side of 0 by plugging into the second derivative. To the left of 0, the value is negative. To the right, the value is positive. Therfore, it is concave up from .

To determine the intervals on which the function is concave down, we must determine the intervals on which the second derivative of the function is negative.

To start, we must find the second derivative:

The derivatives were found using the following rules:

, , ,

Next, we must find the values at which the second derivative is equal to zero:

Using these values, we can make our intervals on which we check the sign of the second derivative:

Next, plug in any point on each interval into the second derivative function and check the sign. On the first interval, the second derivative is negative, on the second, it is positive, and on the third, it is negative. The intervals on which the second derivative is negative are concave down:

.

The trapezoid rule states that 

.

Therefore, using our graph, we have:

We find the function values at the sample point:

Then we substitute the appropriate values into the trapezoid rule approximation:

Is the following function increasing or decreasing at the point ?

Increasing and decreasing intervals can be found via the first derivative. Since derivatives measure rates of change, the sign of the derivative at a given point can tell you whether a function is increasing or decreasing.

Begin by taking the derivative of our function:

Becomes:

Next, find h'(-6) and look at the sign.

So, our first derivative is very negative at the given point. This means that h(x) is decreasing.

The trapezoidal approximation of a definite integral is given by the following formula:

Using the above formula, we get

To evaluate a definite integral using the trapezoidal approximation, we must use the formula

Using the above formula, we get

To evaluate a definite integral using the trapezoidal approximation, we must use the formula

Using the above formula, we get

To evaluate the integral using the trapezoidal rule, we must use the formula

Using the above formula, we get the following: