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 .

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 .

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.

To find when a function is concave, you must first take the 2nd derivative, then set it equal to 0, and then find between which zero values the function is negative.

First, find the 2nd derivative:

Set equal to 0 and solve:

Now test values on all sides of these to find when the function is negative, and therefore decreasing. I will test the values of -3 and 0.

Since the value that is negative is when x=-3, the interval is decreasing on the interval that includes 0. Therefore, our answer is:

This parabola would have the formula . When the first derivative is positive, the function is parabola is increasing. The first derivative is , which is positive on the domain . When the second derivative is positive, the function is concave up. The second derivative is , which is always positive for all real values of .

Therefore, this function is,

Concave up over and increasing over .

To test concavity, we must first find the second derivative of f(x)

This function is concave down anywhere that f''(x)<0, so...

So,

for all

So on the interval -5,-4 f(x) is concave down because f''(x) is negative.

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 .

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

In order to find the intervals of concavity, we must take the second derivative of the function and find the inflection points by setting the setting it to zero (this tells allows us to see when the rate of the rate of change is changing from negative to positive or vice versa). The second derivative is , allowing us to see that our inflection points are . By testing values in between the three different intervals, we can find out which interval is concave down. The interval is concave down because placing any value in between these two numbers into will provide a negative output.

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.