To find the intervals with the same concavity, we need to find the critical points using the second derivative test, then see what the concavity is in the intervals using the second derivative.

; set equal to 0 and solve for , giving  as the only critical point.

Choose an -value either side of the critical point, and test the concavity. For example:

, so the graph is concave down to the left of the critical point.

, so the graph is concave up to the right of the critical point.

Therefore the function is concave down on the interval .

To solve, simply differentiate twice, using the power rule, and then find where the second derivative is negative.

Power rule:

Thus,

Since the second derivative is always negative, our function is concave down everywhere.

To find concavity, you must find the second derivative using the power rule:

Since we are looking for concave down, we are looking for when the second derivative is negative (less than 0). Thus,

Since 14 is always greater than 0, the second derivative is never negative. Thus, our function is never concave down.

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

First, we must find the second derivative:

The derivatives were found using the following rule:

Now, we must find the value at which the second derivative is equal to zero.

We will now use this as the upper and lower limit of our intervals on which we evaluate the sign of the second derivative:

On the first interval, the second derivative is negative, while on the second interval, the second derivative is positive. Thus, our answer is .

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

First, we must find the first and second derivatives:

The derivatives were found using the following rule:

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

Now, we can make the intervals:

Note that at the bounds of the intervals the second derivative is neither positive nor negative.

To determine the sign of the second derivative on the intervals, simply plug in any value on the interval into the second derivative function; on the first interval, the second derivative is positive, on the second it is negative, and on the third it is positive. Thus, the function is concave down on the interval .

Tell whether f(x) is concave up or concave down on the interval [1,2]

To find concave up and concave down, we need to find the second derivative of f(x).

Let's begin by finding f'(x)

Next find f ''(x)

Now, to test for concavity, plug in the endpoints of the interval:

So, on this interval, f"(x) will always be negative. This means that our function is concave down on this interval.

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:

.