Concavity - Online Tutor, Practice Problems & Exam Prep

Determining Concavity from the Graph of f

In previous videos, we learned how the sign of the first derivative of a function shows us whether that function is increasing or decreasing. But what if we took that one step further and actually took a look at the sign of the second derivative? What does that tell us about a function? Well, the sign of the second derivative of a function actually shows us what's called a function's concavity. Now that word concavity may sound a bit intimidating, but all concavity is is just the shape of a graph and how it curves.

Now here, I'm going to walk you through exactly what it means and what it looks like on a graph for a function to be concave up or concave down, and also show you how exactly that relates to the sign of the second derivative. So let's go ahead and get started here. Now let's come right down here to the left side of this graph. We can see that our graph is curving upward. It looks sort of like the smile of a smiley face.

Now whenever we see this shape on a graph, we say that this function is concave up. And when a function is concave up, that corresponds to the second derivative of that function being positive. Now looking over here at the right side of this graph, we can see that it starts to curve downwards. It looks sort of like a frowning face. Now whenever this happens, our function here is concave down, which corresponds to a second derivative of our function being negative.

Now visually, this can be a helpful tool in remembering what it looks like for a function to be concave up or concave down on a graph. Concave up looks like a smile. Concave down looks like a frown. Now even though it's really important to be able to visually recognize the concavity of a function, it's also important to recognize why the second derivative affects that. So let's dive into this more.

Now we know that the first derivative of a function is the rate of change of that function, its slope. So the second derivative of a function is then the rate of change of that first derivative, which we can also think of as being the rate of change of the slope of our function. So let's actually come down back here to this left side of our graph and take a closer look at what's going on with the slope. We can do that by drawing some tangent lines. Now here, we can see that our tangent line starts out with a steep negative slope.

Then it becomes a little bit less steep, still negative, until it becomes 0. So we can see that it starts out very negative and then becomes 0. So it increases from being negative to now being 0. It continues increasing on the other side here, getting pretty steep in the positive direction. So on this whole side that our function is concave up, that slope is increasing.

Now because it's increasing, that means it has a positive rate of change. So that's the exact same thing as saying that it has a positive second derivative. Now, coming back over here to the right side of our function, if we draw some tangent lines over here, we can see that that slope starts out positive, then decreases to 0, continues decreasing as it gets a negative slope. So because this slope is decreasing, it has a negative rate of change and a negative second derivative. So concavity is just how the slope is changing.

That's why it's affected by the sign of our second derivative. Now, having drawn all of these tangent lines on our function here, we can see that on the side where our function is concave up, all of our tangent lines sit underneath the curve. Then where our function is concave down, these tangent lines sit above the curve. This can also be a helpful visual cue in identifying the concavity when given a graph. Now at this point in the middle of this graph, we can see that this is where our function changes concavity from being concave up to now being concave down.

So this is where our function changes concavity, and this point has a special name. It is referred to as an inflection point. Now at an inflection point where our function changes concavity, our second derivative will either be equal to 0 or not exist at all. So similar to a critical point where our first derivative was either equal to 0 or did not exist, this is the same idea, but here for our second derivative, this is an inflection point. So now that we've walked through concavity in detail, let's actually apply this to an example here.

So let's work through this bottom example together. We want to identify where this function is concave up or concave down and also state any points of inflection. Now whenever doing this on a graph, it can be helpful to just go ahead and look at our graph from left to right. So we're going to start on this far left side here from negative infinity. Now immediately looking at this graph, I can see that it's curving downwards.

Remember, concave down looks like a frown. So my function here starts off being concave down from negative infinity. Now where does this interval end? Where does it stop being concave down? Well, continuing on from left to right here, this is where it can be helpful to go ahead and draw some tangent lines.

I can see that our tangent lines are above our curve where it is concave down, then they transition to now being underneath our curve. So this is a cue to me that my function is changing concavity here. So I can see that at this point x equals a negative one, my function changes concavity. That's the end of that interval of being concave down. Now because I know that that's where my function changes concavity, I know that this is a point of inflection, x equals negative one.

So continuing on our function here, at that point negative one begins our next interval where our function is concave up because it looks like a smile there. So our function is in concave up starting from negative one and continuing from left to right here. Where does it stop being concave up? Again, it can be helpful here to go ahead and draw our tangent lines and pay attention to what's going on with the slope here. We can see that these tangent lines transition from being underneath our curve to now being on top right here at x equals 2.

So this indicates to me yet another point of inflection, x equals 2, and the end of that interval of my function at being concave up. Then it starts to curve downwards again, concave down, looks like a frown, and that continues on to infinity. So my function is again concave down from 2 on to infinity. So here we have all of our intervals of concavity along with our inflection points. In the next couple of videos, we're going to continue learning more about concavity and getting more practice.

I'll see you there.

For the following graph, find the open intervals for which the function is concave up or concave down. Identify any inflection points.

• A. Concave up: $\left(-\infty,0\right)$, $\left(1,\infty\right)$; Concave down: $\left(0,1\right)$; Inflection points: $\left(0,0\right)$, $\left(1,1\right)$
• B. Concave up: $\left(-\infty,0\right)$; Concave down: $\left(0,\infty\right)$; Inflection points: $\left(0,0\right)$
• C. Concave down: $\left(-1,0\right)$, $\left(1,\infty\right)$; Concave up: $\left(0,1\right)$; Inflection points: $\left(0,0\right)$, $\left(1,1\right)$
• D. Concave up: $\left(-1,0\right)$, $\left(1,\infty\right)$; Concave down: $\left(0,1\right)$; Inflection points: $\left(-1,1\right)$, $\left(0,0\right)$, $\left(1,1\right)$

For the following graph, find the open intervals for which the function is concave up or concave down. Identify any inflection points.

• A. Concave up: $\left(-2,-1\right)$, $\left(2,4\right)$; Concave down: $\left(-1,1\right)$, $\left(1,2\right)$, $\left(4,5\right)$; Inflection points: $\left(-1,0\right)$, $\left(1,0\right)$, $\left(2,-1\right)$, $\left(4,3\right)$
• B. Concave up: $\left(-2,-1\right)$, $\left(2,4\right)$; Concave down: $\left(-1,2\right)$, $\left(4,5\right)$; Inflection points: $\left(0,1\right)$, $\left(1,0\right)$, $\left(2,-1\right)$, $\left(4,3\right)$
• C. Concave up: $\left(-2,1\right)$, $\left(2,4\right)$; Concave down: $\left(1,2\right)$, $\left(4,5\right)$; Inflection points: $\left(1,0\right)$, $\left(2,-1\right)$, $\left(4,3\right)$
• D. Concave up: $\left(-2,0\right)$$\left(-2,1\right)$, $\left(2,4\right)$; Concave down: $\left(0,2\right)$, $\left(4,5\right)$; Inflection points: $\left(0,1\right)$, $\left(1,0\right)$, $\left(2,-1\right)$, $\left(4,3\right)$

For the following graph, find the open intervals for which the function is concave up or concave down. Identify any inflection points.

• A. Concave down: $\left(0,\infty\right)$; No inflection pt
• B. Concave down: $\left(0,\infty\right)$; Inflection pt: $\left(0,0\right)$
• C. Concave up: $\left(-\infty,0\right)$; Concave down: $\left(0,\infty\right)$; Inflection pt: $\left(0,0\right)$
• D. Concave down: $\left(-\infty,\infty\right)$; No inflection pt

Determining Concavity from the Graph of f' or f''

Hey everyone. We just learned how to identify the concavity of a function by looking at its graph. And here for this function, we can see that it is concave up on this left side and concave down on this right side with an inflection point where it changes concavity right here in the middle. Now, we can tell all of that information by just looking at the shape of this graph, but we won't always be given the graph of a function itself. We may be given the graph of that function's derivative or even the graph of that function's second derivative.

And we'll still need to be able to identify the concavity of the actual function. Now this may sound like it's going to be complicated, but we already know everything that we need to in order to do this. It's just about translating what we already know to these graphs, which I'm going to help you out with here. So let's go ahead and get started. Now in the previous video, we talked about how the sign of the second derivative tells us the concavity of a function.

If that second derivative is positive, then our function is concave up. If that second derivative is negative, our function is concave down. So when given the graph of a function's second derivative, how does that translate? Well, looking at the graph here that we have of our function's second derivative, we can see that on this left side, this graph is all positive.

It's all above the x-axis, which means that this second derivative is positive, corresponding to our original function being concave up. Then on this right side here, we can see that our graph becomes negative. It's all below the x-axis. So here, our second derivative is negative, corresponding to our original function of being concave down. So when given the graph of a function's second derivative, we just need to pay attention to whether that graph is above or below the x-axis, whether it's positive or negative.

If it's positive, then our original function is concave up, and if it's negative, our original function is concave down. We can also locate our inflection point here. We can see that where this graph changes sign right here in the middle, this would be the location of our inflection point. Now, because here we are just working with the graph of our second derivative, it would just be the x value of that inflection point and not the entire ordered pair.

Let's make our way back to our first derivative. Now in the previous video, we also talked about our second derivative being the rate of change of our first derivative. Now because of that, we can determine concavity by whether that first derivative is increasing or decreasing, which is something that we can easily identify on a graph. So looking at the graph of our function's first derivative here, we can see that on this left side, this graph is going up. It is increasing, which corresponds to our original function being concave up.

Then on this right side here, our graph is falling back down. It is decreasing, which corresponds to my original function being concave down. Now again, here I can go ahead and locate my inflection point where this graph changes from increasing to decreasing or would also be where it changes from decreasing to increasing. Now again, just like for the graph of our second derivative, this would just be the x value of that inflection point and not the entire ordered pair. So now that we've seen both graphs of our first derivative and our second derivative, let's put all of this information together.

If our original function f is concave up, our first derivative will be increasing, and our second derivative will be positive. Both things that we can identify by looking at their graphs. Now if our original function is concave down, our first derivative will be decreasing and our second derivative will be negative. So when you're given a graph and asked to determine concavity, make sure you know what graph you're given; the graph of a function, the graph of a function's derivative, or the graph of a function's second derivative. Now that we know how to identify this by looking at all of these graphs, let's get a bit more practice. I'll see you in the next video.

The graph of $f^{\prime}\left(x\right)$ is shown below. Use the graph to determine the intervals for which $f\left(x\right)$ is concave up or concave down and the location of any inflection points.

• A. Concave up: $\left(-\infty,0\right)$, $\left(1,\infty\right)$; Concave down: $\left(0,1\right)$; Inflection points: $x=0$, $x=1$
• B. Concave up: $\left(-\infty,0\right)$, $\left(1,\infty\right)$; Concave down: $\left(0,1\right)$; Inflection points: $x=-1$, $x=0$, $x=1$
• C. Concave down: $\left(-1,0\right)$ , $\left(1,\infty\right)$; Concave up: $\left(0,1\right)$; Inflection points: $x=0$, $x=1$
• D. Concave down: $\left(-1,0\right)$ , $\left(1,\infty\right)$; Concave up: $\left(0,1\right)$; Inflection points: $x=-1$, $x=0$, $x=1$

The graph of $f''(x)$ is shown below. Use the graph to determine the intervals for which $f\left(x\right)$ is concave up or concave down and the location of any inflection points.

• A. Concave up: $\left(-1,0\right)$, $\left(2,\infty\right)$; Concave down: $\left(0,2\right)$; Inflection points: $x=0$, $x=2$
• B. Concave up: $\left(-1,0\right)$, $\left(2,\infty\right)$; Concave down: $\left(0,2\right)$; Inflection points: $x=-1$, $x=0$, $x=2$
• C. Concave up: $\left(-\infty,0\right)$, $\left(2,\infty\right)$; Concave down: $\left(0,2\right)$; Inflection points: $x=0$, $x=2$
• D. Concave up: $\left(-\infty,-1\right)$, $\left(0,\infty\right)$; Concave down: $\left(-1,0\right)$; Inflection points: $x=-1$, $x=0$

Determining Concavity from the Graph of f' or f'' Example 1

In this problem, we're given the graph of a function and asked to identify where this function is increasing and decreasing, where this function is concave up or concave down, and the coordinates of any points of inflection. This is a lot of information, but we can break it down. Remember, a function is increasing or decreasing where it's either going up or going down. It's concave up if it appears like a smile, concave down if it appears more like a frown. Points of inflection occur where the concavity changes.

Let's start by examining points a and b and identifying where our function is increasing or decreasing. Observing the function from left to right, beginning from the endpoint at negative 5, we see the function is decreasing. It starts off decreasing from negative 5 and continues until it reaches negative 4. So, in this short interval from negative 5 to negative 4, the function is decreasing.

Then, it begins to increase all the way until it reaches negative 2. This interval of increase extends from negative 4 to negative 2. Then, the function starts decreasing again from negative 2 and continues until it reaches positive 1. Therefore, the function is decreasing again from negative 2 up to positive 1.

Upon reaching positive 1, the function begins to increase and continues until it hits 3. From this point at 1 to 3, the function continues increasing. It then decreases again from 3 to 5. Towards the end, it picks up again, increasing from 5 until the endpoint at 6, thus covering all our intervals of increasing and decreasing. Now, let's examine the concavity of our function.

In general, concave up will resemble a smile and concave down will look like a frown. We also consider our tangent lines: if a tangent line lies underneath the curve, the function will be concave up. If the tangent line lies on top of the curve, it will be concave down. Starting at the left endpoint at negative 5, the function appears like a smile, thus indicating it is concave up from negative 5 until negative 3. Confirmation of this is seen by the tangent lines transitioning from being underneath to being on top. So, from negative 3, the function appears like a frown, indicating concave down, and this continues until x equals 0.

Observing the shape of our function, we see it transitions from curving downwards to upwards around x equals 0, signaling a change in concavity. Thus, from 0 to 2, the function is concave up until we see another transition at 2. From 2 to 4, the function is concave down, and from 4 to 6, it is concave up again, defining our intervals of concavity.

Additionally, the coordinates of our points of inflection, where our concavity changes, are: at x equals negative 3, the coordinate is (-3, 1); at x equals 0, the coordinate is (0, -2); at x equals 2, the coordinate is (2, -3); and at x equals 4, the coordinate is (4, -3). These points lie between the intervals where the concavity shifts from up to down, or vice versa.

It's important to note that increasing or decreasing statuses of a function don't necessarily correlate with its concavity. These concepts are derived from the first and second derivatives respectively. This understanding is crucial as it separates the increasing or decreasing nature of a function from whether it is concave up or down. The first derivative reveals if a function is increasing (positive derivative) or decreasing (negative derivative), while the second derivative determines its concavity: concave up if positive and concave down if negative.

If you have any questions, please feel free to ask, and let's continue learning!

Determining Concavity Given a Function

So far, we've learned a ton of different information about concavity, and we know that the sign of the second derivative of a function tells us whether that function is concave up or concave down. Now we also know that concavity changes at what's called an inflection point, and at this inflection point, our second derivative will either be equal to 0 or not exist. Now everything that we've learned so far with concavity has been in relation to looking at graphs, but we won't always be given the graph of a function, and we'll still need to be able to determine where a function is concave up or down just based on the function itself. And that's exactly what I'm going to show you how to do here using all of the information that we already know about concavity. And I think you'll find this to be a really familiar process to things we've done before.

So let's go ahead and get started here. Now, in order to determine where a function is concave up or concave down, all we have to do is find the potential inflection points of a function. That's going to be where our second derivative is either equal to 0 or does not exist. And then find the sign of our second derivative, testing that between these potential inflection points. Now, I have a step-by-step process that we can follow in order to do this.

And I think you'll find this a really familiar process to how we determine the intervals where a function was increasing or decreasing. This is almost identical, except here we're going to be using the second derivative since that tells us concavity as opposed to the first. So let's go ahead and jump into our example here. Here, we're asked to determine the intervals for which our function f(x) is concave up or concave down. Now the function that we have here is \( f(x) = 2x^3 + 12x^2 + 9x - 4 \).

And we wanna start here by finding our potential inflection points. Remember, like I said earlier, this is where that second derivative is either equal to 0 or it does not exist. So in order to find these points, we first need to find that second derivative, which, of course, starts with finding our first derivative. So finding that first derivative here using the power rule, this gives me \( 6x^2 + 24x \). Then if I differentiate again here getting my second derivative, \( f''(x) \), that's going to be \( 12x + 24 \), or I can factor a 12 out of both of these terms to give me \( 12(x + 2) \).

So now that I have that second derivative, I know that there's nowhere the second derivative won't exist, so I can find my potential inflection points here by setting this equal to 0. Now, doing that, this 12 will cancel, and I just need to subtract 2 from both sides to give me that inflection point at \( x = -2 \). Now you'll notice that this step specifically refers to these as potential inflection points. And the reason for that is we don't actually know if the concavity changes at this point yet, and we won't know until we get to our final step here. So this just could be an inflection point.

We don't know that it definitely is yet. But now that we have this potential inflection point, let's move on to our second step here. This is where we wanna make our sine chart intervals, but here based on those potential inflection points. Since we just have the 1, we can go ahead and mark that negative 2 on our sign chart here, and this splits our sign chart into 2 intervals from negative infinity to negative 2 and from negative 2 on to infinity. So I have my intervals there, and now we can move on to our last step where we wanna choose a test value in each of these intervals to plug back into our second derivative.

So when choosing test values, we've done this before. Remember that we wanna choose them so that they're easy to plug back into whatever we're plugging into here, our second derivative. So, in this interval from negative infinity to negative 2, I'm gonna go ahead and choose a negative 3, and then from negative 2 to infinity, I'm just going to choose 0. So I wanna plug these specifically into that second derivative. So \( f''(-3) \) and \( f''(0) \).

Now starting with negative 3 here, plugging this into the factored version of that second derivative, that's going to be \( 12(-3 + 2) \) which ends up giving me a negative 12. Now here, we don't actually care about the value, and we've seen this before because we just care about the sign. Now here, our second derivative is negative. So what does that tell us about our function? Remember that wherever our function's second derivative is negative, that means that our function is concave down.

And here, that tells us that it's concave down on this entire interval, the same way that when we found our first derivative was decreasing or increasing over an entire interval. This is the same idea here now for concavity using our second derivative. So based on our second derivative here being negative for our test value, we now know that our function is concave down from negative infinity all the way to negative 2. Now let's go ahead and work with our second test value here in our other interval. Plugging this into that second derivative, that's \( 12(0 + 2) \), this results in a positive 24.

Again, we don't care about what that value is, just what its sign is, and here it is positive. So here, since that second derivative is positive, we know that that means that our function is concave up here on this entire interval. So our function here is concave up from negative two to infinity. So here we have our intervals of concavity that we determined based on the sign of our second derivative. We know that concave down looks like a frown, concave up looks like a smile, so we can kind of envision what our graph might look like.

Now earlier, we said that negative 2 was a potential inflection point. And here, based on what we found about the concavity of our function, it changes from being concave down to then being concave up. Because our concavity changes at this point, we know that it is indeed an inflection point. Now, we're going to get some more practice with concavity coming up next. So I'll see you in the next one.

Determining Concavity Given A Function Example 2

In this problem, we want to determine the intervals for which this function is concave up or concave down and also state any inflection points. Remember that a function's concavity is based on the sign of the second derivative. If the second derivative is positive, our function is concave up there. And if the second derivative is negative, our function will be concave down. Now inflection points are just where that concavity changes, so either from concave up to concave down or vice versa.

Now the function that we're working with here is f(x) equals 9 times x to the power of 1/5, so we're dealing with a rational exponent here. Now the very first thing we want to do is find any potential inflection points of our function. That's where the second derivative is either equal to 0 or does not exist. So we need to start here by just finding that second derivative, which we, of course, have to find our first derivative before we can get to our second. So our first derivative here, f'(x), using the power rule, pulling that 1/5 out to the front, multiplying it by that 9 gives me 9/5.

Times x to the power of 1/5 minus 1, that gives me negative 4/5. Now I need to find my second derivative here, differentiating again, pulling that negative 4/5 out to the front, multiplying it by that 9/5 will give me negative 36 over 25. And then times x to the power of negative 4/5 minus 1, that gives me negative 9/5. Now remember that whenever we have a negative exponent, that just means that we can rewrite this to be on the bottom of a fraction. So this is negative 36 over 25 times x to the power of positive 9/5.

So putting it on the bottom of our fraction there, I can now make that exponent positive. So now that we have our second derivative, we want to find those potential inflection points by setting this equal to 0 and then determining where it does not exist. Now because we don't have any variables in our numerator here, there's not going to be anywhere where this second derivative is equal to 0. So I instead need to focus on where it does not exist. Now because my denominator here has that x in it, I can set this equal to 0 to determine where this doesn't exist.

And because I have this 25, I'm multiplying this x to the power of 9/5. This will specifically happen when x is equal to 0. So here is where our second derivative does not exist at x equals 0. So this is our only potential inflection point that we can now use to make our sign chart intervals. So setting up our sign chart here, marking our only potential inflection point, that's x equals 0, splits my sign chart into 2 intervals, from negative infinity to 0 and from 0 to infinity.

Now with that sign chart made, I want to go ahead and plug a test value from each interval into my second derivative here. Now I know that throughout this chapter, we've been working with the function, we've been working with the function's derivative, and now its second derivative. So be really careful here and make sure that you're plugging things into the correct thing. Now here, we're going to be plugging test values specifically into that second derivative. So let's choose our test values here.

Remember, we want to choose them just so that it's easy for us to plug back in. Now on this interval from negative infinity to 0, I'm going to go ahead and choose negative 1. Then from 0 to infinity, I will choose positive 1. Now we're plugging these specifically back into that second derivative. That's f''(x).

So let's go ahead and do that here, f''(-1) and f''(1). Plugging in that negative one, that gives me -36/25 times (-1) to the power of 9/5. Now remember, we only care about the sign of this value. So you can choose to fully calculate this to get the whole value itself, or you can just pay attention to what's going on with the sign. Here, I have a negative value.

Then I'm taking this negative one, raising it to the power of 9/5. That will still result in a negative. So I have a negative being divided by a negative. This results in a positive. Now if you actually calculated out this value, you'll find that this gives you positive 36/25. But we really only care that this is positive. Because this value is positive, that tells us that our function is concave up on this entire interval. So based on that positive second derivative here, that's positive, our function is concave up on the interval from negative infinity to 0.

Now let's go ahead and plug our other test value in here, 1. That'll be -36/25 times 1 to the power of 9/5. That's a positive one here. Now seeing that this is a negative and then on the bottom, all I have is positives, I can already tell this is going to be a negative value. If you were to work out that value, you'd get -36/25. Again, we don't care what that value is, just that it is negative. Because this value is negative, this tells us that our function is going to be concave down on this interval.

So this negative second derivative leads me to know that this function is concave down from 0 to infinity. So we have some information here about our intervals of concavity. This function is concave up from negative infinity to 0 and concave down from 0 to infinity. Now we can see at this point, x equals 0, that our function changed from being concave up to then being concave down. Now because of that, x equals 0 represents an inflection point.

So I have an inflection point at x equals 0. Now if you want to go ahead and find exactly where this inflection point is located, you can plug this back into your original function. So if I want to find where this inflection point is located, I can find f(0), plugging in that x value there. This will give me 9 times 0 to the power of 1/5, which is also just 0. So my inflection point as an ordered pair is located at (0,0).

Here we found our intervals of concavity as well as our one inflection point. Feel free to let us know if you have any questions here, and we'll continue practicing in the next one.

Determining Concavity Given a Function Example 3

In this problem, we're asked to determine the intervals for which this function is concave up or concave down and also state any inflection points. Remember that a function is concave up where its second derivative is positive and concave down where its second derivative is negative. Inflection points are anywhere that the concavity changes. Here, the function we're working with is a polynomial function, specifically f(x)=5x2−2x+24.

To determine the intervals of a function's concavity, we first want to find any potential inflection points where the second derivative is either equal to zero or does not exist. We start by finding the first derivative. Using the power rule, we obtain f′(x)=10x−2. Differentiating again, the second derivative is f′′(x)=10.

When finding potential inflection points, we check where the second derivative is either zero or does not exist. However, in this case, the second derivative is a constant positive 10. Hence, it will neither be zero nor undefined at any point. Since our second derivative is positive throughout, the function is concave up over the entire real number line, from negative infinity to positive infinity.

Given the nature of the quadratic function 5x2−2x+24, which is an upward-facing parabola, we can visually confirm it resembles a smile shape, indicating concavity upwards throughout. Since the concavity never changes, there are no inflection points.

Feel free to let me know if you have any questions here. I'll see you in the next one.

Determine the intervals for which the function is concave up or concave down. State the inflection points.

$f(x)=2x(x+9)^2$

• A. Concave down: $\left(-\infty,\infty\right)$ ; No Inflection Points
• B. Concave down: $\left(-\infty,-6\right)$; Concave up: $\left(-6,\infty\right)$; Inflection pt: $\left(-6,-108\right)$
• C. Concave down: $\left(-\infty,6\right)$; Concave up: $\left(6,\infty\right)$; Inflection pt: $\left(6,2700\right)$
• D. Concave down: $\left(-\infty,0\right)$; Concave up: $\left(0,\infty\right)$; Inflection pt: $\left(0,0\right)$

Determine the intervals for which the function is concave up or concave down. State the inflection points.

$f(x)=\frac{7}{2x-9}$

• A. Concave down: $\left(-\infty,\infty\right)$; No Inflection Points
• B. Concave up: $\left(-\infty,\infty\right)$; No Inflection Points
• C. Concave down: $\left(-\infty,\frac92\right)$ ; Concave up: $\left(\frac92,\infty\right)$; Inflection point: $\left(\frac92,0\right)$
• D. Concave down: $\left(-\infty,\frac92\right)$ ; Concave up: $\left(\frac92,\infty\right)$; No Inflection Points

Determine the intervals for which the function is concave up or concave down. State the inflection points.

$f(x)=4\ln(3x^2)$

• A. Concave down: $\left(-\infty,0\right)$; Concave up: $\left(0,\infty\right)$; Inflection pt: $\left(0,0\right)$
• B. Concave down: $\left(-\infty,0\right)$ & $\left(0,\infty\right)$; No inflection pt
• C. Concave up: $\left(-\infty,0\right)$; Concave down: $\left(0,\infty\right)$; Inflection pt: $\left(0,0\right)$
• D. Concave up: $\left(-\infty,0\right)$; Concave down: $\left(0,\infty\right)$; No inflection pt