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 gonna 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: $\left(-2,0\right)$, $\left(2,4\right)$; 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: $\left(0,\frac{\pi}{2}\right)$; Concave up: $\left(\frac{\pi}{2},\pi\right)$; Inflection pt: $\left(\frac{\pi}{2},0\right)$
• B. Concave down: $\left(0,\frac{\pi}{2}\right)$; Concave up: $\left(\frac{\pi}{2},\pi\right)$; Inflection pt: $\left(0,\frac{\pi}{2}\right)$
• C. Concave down: $\left(0,\frac{\pi}{2}\right)$; Concave up: $\left(\frac{\pi}{2},\pi\right)$; Inflection pts: $\left(\frac{\pi}{4},5\right)$, $\left(\frac{3\pi}{4},-5\right)$
• D. Concave down: $\left(\frac{\pi}{4},\frac{3\pi}{4}\right)$ ; Concave up: $\left(0,\frac{\pi}{4}\right)$, $\left(\frac{3\pi}{4},\pi\right)$; Inflection pts: $\left(\frac{\pi}{4},5\right)$, $\left(\frac{3\pi}{4},-5\right)$

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 actually 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 actually 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 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. So that's our second derivative.

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 down: $\left(-1,0\right)$, $\left(2,\infty\right)$; Concave up: $\left(0,2\right)$; Inflection points: $x=0$, $x=2$
• B. Concave down: $\left(-1,0\right)$, $\left(2,\infty\right)$; Concave up: $\left(0,2\right)$; 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 the following: where this function is increasing and decreasing, where this function is concave up or concave down, and the coordinates of any points of inflection. Now, this is a lot of information, but we can break this down. Remember that a function is increasing or decreasing where it's either going up or going down. It's concave up if it looks somewhat like a smile, concave down if it looks sort of like a frown. Then any points of inflection are where our concavity changes.

So let's start here by taking a look at a and b and identifying where our function is increasing or decreasing. Now we're going to be looking at this function from left to right. Starting from that endpoint negative 5, we can see that our function is going down. So it starts off decreasing from negative 5 and keeps going down until we reach right here at negative 4. So that's a rather short interval from negative 5 to negative 4 where this function is decreasing.

Then it begins going up and increasing all the way until we reach x equals negative 2 right here. So that second interval where my function is increasing goes from negative 4 to negative 2. Then continuing on our function here, it starts going down from a negative 2. It continues going down until we reach positive 1 here. So our function is again decreasing from negative 2 all the way to positive 1.

When we reach positive 1, it starts going up again and then continues going up until we reach right here at 3. So from this point at 1 until we reach 3, our function continues increasing. Then it drops back down, decreasing from 3 to 5. Then, finally, it continues going up from 5 until our other endpoint at 6 so we have all of our intervals of increasing and decreasing. Now let's look at parts c and d and determine the concavity of our function.

Remember that in general, concave up will look like a smile, and concave down will look like a frown. We can also think about this in terms of tangent lines. If we draw a tangent line to our function and it is underneath our curve, we know that it will be concave up. If we draw our tangent line and it is on top of our curve, it will be concave down. Starting at our function at the first endpoint, negative 5, we can see that it looks like a smile here.

So, from negative 5, we are curving upward looking like a smile, and then it continues to be concave up until we reach about negative 3. We confirm this by drawing our tangent lines here. We see that's where it transitions from being underneath to being on top. So our function starts out at concave up from that endpoint negative 5 until we reach negative 3. Then from negative 3, it curves the other way.

It looks like a frown concave down from negative 3 for quite a while continues being concave down until x equals 0 when we see those tangent lines go from being on top of our curve to now being underneath our curve. It can be a little bit tricky to sometimes identify where our concavity changes. But as we look at the curve of our function, we can see that it transitions from curving downwards to curving upwards, and that's where our concavity changes. From negative 3 until we reach x equals 0, our function is concave down. Then it begins to be concave up again from 0, curving up, until we reach positive 2, where we can again see that transition with our tangent lines.

It then starts to curve downwards. So from 0 to 2, concave up begins to be concave down again from 2 until we reach 4. So from 2 to 4, concave down again. We can again see that we have our transitioning tangent lines from being on top of our function to now being underneath our function. And from 4 on to that endpoint 6, our function is concave up. And that wraps up our intervals of concavity. Now we have one final question here. And that's the coordinates, the x and y coordinates, of our points of inflection. These points are in between all of these intervals where our concavity changed from being up to down or being down to up.

Our points of inflection are at (-3, 1), (0, -2), (2, -3), and (4, -3). These points are where our concavity changes. Remember, these are based on the second derivative. Just because a function is increasing, that does not mean that a function is concave up. These are completely different things, one of them based on the first derivative and the other on the second derivative. So it's important to separate these concepts in our minds. Just because a function is increasing doesn't mean it's concave up, and just because a function is decreasing doesn't mean it's concave down.

If you have any questions, feel free to let us know and let's keep 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 \( f(x) = 2x^3 + 12x^2 + 9x - 4 \) is concave up or concave down.

And we want to 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, \(\)\) - that's going to be \(\)\), or I can go ahead and 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 want to make our sign chart intervals, but here based on those potential inflection points. Since we just have the one, 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 want to 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 want to 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 going to go ahead and choose a negative 3, and then from negative 2 to infinity, I'm just going to choose 0. So I want to plug these specifically into that second derivative. So \(\)\) here of a negative 3 and \(\)\) of 0.

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 that second derivative is negative, our function will be concave down. Now, inflection points are just where that concavity changes, either from concave up to concave down or vice versa.

Now the function that we're working with here is \( f(x) = 9 \times x^{\frac{1}{5}} \), 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 that 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 \( \frac{1}{5} \) out to the front, multiplying it by that 9 gives me \( \frac{9}{5} \).

Times \( x^{\frac{1}{5} - 1} \), that gives me \( x^{-\frac{4}{5}} \). Now I need to find my second derivative here, differentiating again, pulling that \( -\frac{4}{5} \) out to the front, multiplying it by that \( \frac{9}{5} \) will give me \( -\frac{36}{25} \). And then times \( x^{-\frac{4}{5} - 1} \), that gives me \( x^{-\frac{9}{5}} \). 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 \( -\frac{36}{25} \times x^{\frac{9}{5}} \).

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 \( x^{\frac{9}{5}} \). This will specifically happen when \( x \) is equal to 0. So here is where our second derivative does not exist, at \( x = 0 \). 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 \( x = 0 \), 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 \( -\frac{36}{25} \times (-1)^{\frac{9}{5}} \). 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 \( \frac{9}{5} \). 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 will find that this gives you positive \( \frac{36}{25} \).

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 in our other test value here, 1. That'll be \( -\frac{36}{25} \times (1)^{\frac{9}{5}} \).

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 \( -\frac{36}{25} \). 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, \( x = 0 \), that our function changed from being concave up to then being concave down. Now because of that, \( x = 0 \) represents an inflection point.

So I have an inflection point at \( x = 0 \). 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 \( 9 \times (0)^{\frac{1}{5}} \), 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 occur where the concavity changes. The function we're working with is a polynomial function, given by fx=5x2-2x+24.

To determine the intervals of a function's concavity, we first need to find any potential inflection points, where the second derivative is either zero or does not exist. To do this, we start by finding the first derivative: f'x=10x-2, using the power rule. Differentiating again, the second derivative f''x=10 is a constant.

When looking for potential inflection points by finding where the second derivative is either zero or nonexistent, we note that our second derivative is uniformly +10. Since 10 does not equal zero and does not cease to exist, our function has a constantly positive second derivative. Therefore, our function is concave up from negative infinity to positive infinity. This condition confirms that the function is a quadratic and specifically an upward-facing parabola, as expected with a positive coefficient on the x2 term. Consequently, it resembles a smile and is always concave up.

Because the concavity never changes across its domain, there are no inflection points. This function has no points where the concavity changes. Please let me know if you have any questions. 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. 

$f(x)=2\sin x+3x$; $0$ < $x$ < $2\pi$

• A. Concave down: $\left(0,\pi\right)$; concave up: $\left(\pi,2\pi\right)$; Inflection pt: $\left(\pi,3\pi\right)$
• B. Concave down: $\left(0,\pi\right)$; concave up: $\left(\pi,2\pi\right)$; Inflection pt: $\left(\pi,2+3\pi\right)$
• C. Concave down: $\left(0,\frac{\pi}{2}\right)$, $\left(\pi,\frac{3\pi}{2}\right)$; concave up: $\left(\frac{\pi}{2},\pi\right)$, $\left(\frac{3\pi}{2},2\pi\right)$; Inflection pts: $\left(\frac{\pi}{2},\frac{4+3\pi}{2}\right)$, $\left(\pi,3\pi\right)$, $\left(\frac{3\pi}{2},\frac{-4+9\pi}{2}\right)$
• D. Concave up: $\left(0,2\pi\right)$; No inflection pt