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

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

The derivatives were found using the following rules:

, , , 

Next, we find the values on the given interval for which the second derivative is equal to zero:

We now use these values as bounds for the intervals on which we check the sign of the second derivative:

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

Evaluating the sign simply by plugging in any value on the given interval into the second derivative function, we find that on the first interval, the second derivative is negative, while on the second interval, the second derivative is positive. Thus, the function is concave up on the second interval, .

Tell whether g(w) is concave up or concave down when . How do you know?

To test for concavity, we need to know the sign of our function's second derivative at the given value.

Let's begin by finding the first derivative. To do so, we need to recall the following:

Those two rules tell us how to derive functions made up of polynomials and exponential terms.

Apply the rules to get the following.

Now, do derive again to get the 2nd derivative.

Now, we are almost there, but we need to find g"(-3)

So, our second derivative is positive when w=-3. This means our original function is concave up at this point.

Concave up, because 

The function is concave down wherever , so we compute  and see where it is negative. We have:

 (a parabola, opening upwards)

To find where  is negative, we first find its zeros by setting :

,

so  when  or , and we conclude that  is negative ( is concave down) between them. That is, . The only answer choice completely inside this interval (not outside or at the endpoints) is

.

Even though the first derivative () is  at , there is no max or min because the function is increasing on both sides (derivative is positive on both sides). However, the function is changing its concavity (from negative to positive) at . We can tell because the second derivative () is also  at , and it's going from negative to positive. Hole and asymptote are irrelevant. When the second derivative changes signs around a specific point we call this an inflection point. An inflection point describes a point that changes the concavity of a function.

Since the function  is increasing at , , and since  is below the x-axis, .

Furthermore, there exists an inflection point at , where the concavity of function changes.

Thus at , .

Therefore, the correct answer is . 

Inflection points can only occur when the second derivative is zero or undefined. Here we have

. 

Therefore possible inflection points occur at  and . However, to have an inflection point we must check that the sign of the second derivative is different on each side of the point. Here we have

. 

Hence, both are inflection points

Possible inflection points occur when  . This occurs at three values, . However, to be an inflection point the sign of  must be different on either side of the critical value. Hence, only  are critical points.

A point of inflection is found where the graph (or image) of a function changes concavity. To find this algebraically, we want to find where the second derivative of the function changes sign, from negative to positive, or vice-versa. So, we find the second derivative of the given function  

The first derivative using the power rule  

 is,

 and the seconds derivative is  

We then find where this second derivative equals .   when .

We then look to see if the second derivative changes signs at this point. Both graphically and algebraically, we can see that the function  does indeed change sign at, and only at, , so this is our inflection point.

Infelction points are the points of a graph where the concavity of the graph changes.  The inflection points of a graph are found by taking the double derivative of the graph equation, setting it equal to zero, then solving for .  

To take the derivative of this equation, we must use the power rule,

 .  

We also must remember that the derivative of an constant is 0.  

After taking the first derivative of the graph equation using the power rule, the equation becomes

.  

In this problem the double derivative of the graph equation comes out to , factoring this equation out it becomes .  

Solving for when the equation is set equal to zero, the inflection points are located at .

In order to find the points of inflection, we need to find  using the power rule .

Now to find the points of inflection, we need to set .

.

Now we can use the quadratic equation.

Recall that the quadratic equation is

,

where a,b,c refer to the coefficients of the equation  .

In this case, a=12, b=0, c=-4.

Thus the possible points of infection are

.

Now to check if or which are inflection points we need to plug in a value higher and lower than each point. If there is a sign change then the point is an inflection point.

To check  lets plug in .

Therefore  is an inflection point. 

Now lets check  with .

Therefore  is also an inflection point.