Hey, everyone. So up to this point, we've spent a lot of time talking about graphs. And in this video, we're going to see if we can apply this concept of graphs to this new topic of relations and functions. Now, this topic is often considered confusing when students initially encounter it. But throughout this video, we're going to be going over a lot of different scenarios and examples to see if we can really clear up some of this confusion around this subject. So let's get right into this. Relations are a connection between X and Y values and graphically they are represented as ordered pairs. Now functions are a special kind of relation where each input has at most one output and it's important to note that all functions are relations, but not all relations are functions. So to understand this a little bit better let's take a look at this example that we have down here. We have these two graphs of relations and we want to see if we can determine whether these are also examples of functions. We'll start with this graph on the left, and what I'm going to do is write out all of the inputs which correspond to all of the X values. So I'll do this in ascending order so first I see that we have an X value of negative 2. I also see that we have an X value of positive one and even though it shows up twice here it's perfectly okay to only write it once in this bubble down here. Now lastly I see that we have an X value of 3. So these are all of the inputs that we have. Now the outputs are going to correspond to the Y values, and I'll list all of these out as well. So I see that we have positive 2, I see that we have 4, we also have 1, and then we have negative 2. So these are all of the inputs and outputs. Now looking at these points I see that negative 2 is related to positive 2. I see that positive one is related to 4, and I also see that one is related to 1. Now lastly I see that we have this point which says 3 is related to negative 2. Now based off this relation that we see, can we conclude whether or not it's a function? Well, recall that we set up here for a relation to be a function, each input can have at most one output. And if I look at each of our inputs, I can see that there is an input that has more than one output. And as soon as this happens, you can automatically conclude that this is not an example of a function. But let's take a look at this other example for this graph on the right. So what I'm gonna do with this graph is I'm going to list out all of the inputs in ascending order like we did before. So this will be all the X values, so I see that we have negative 4, I see that we have negative 2, I see that we have 1, and then we have 3. Now what I'm also going to do is list out all of the outputs, which correspond to all the Y values. So I see that we have positive 2, I see that we have negative 1, and I also see that we have positive 2 up here, but since we already wrote positive 2 once we don't have to write it again. So we have positive 2, and then we have positive 4. So these are all of the outputs. Now looking at how these are related I can see that negative 4 is related to negative or to excuse me to positive 2. I see that negative 2 is related to negative 1, and I see that positive 1 is related to positive 2, and then I see that we have that positive 3 is related to positive 4. Now given this information, can we conclude whether or not this is a function? Well, we need to see if any of the inputs have more than 1 output, and if I look at this each of these inputs only go to one output, and because of this, we would say this is an example of a function. So this is how you can tell whether or not a relation is a function, but you may have noticed this process was a bit tedious, having to write out all the inputs(outputs like this. Well, you may be happy to know there is a shortcut to solving these problems, and the shortcut is called the vertical line test. This states that if you can draw any vertical line that passes through more than one point on your graph, then the graph is not going to be a function. So let's try this vertical line test on the two graphs we had up here. I'll take vertical lines and I'll draw them through every point that I see on the graph, and let me draw this vertical line a bit better. So if I draw these vertical lines, I notice that there is a place where the vertical line passes through more than one point. If this ever happens, then it's not a function. But let's try the vertical line test on this other graph. If I draw vertical lines through each of the points that I see, I noticed that no matter where I draw a vertical line, I'm only ever gonna pass through one point at most. And because of this, we would say that this is an example of a function. Now let's take a look at a couple more examples, because you could also use the vertical line tests on graphs like this. If I tried drawing some vertical lines for this graph on the left, I noticed that we will have some vertical lines that pass through more than one point. And because of this, we can conclude that this is not an example of a function. But if I try the same vertical line test on the other graph that we have over here, notice that no matter where I draw a vertical line, we're only ever going to pass through one point at most, and because of this, we can conclude that this graph is an example of a function. So that's the basic idea of relations and functions. Hopefully, this helped you out and let me know if you have any questions.

# Intro to Functions & Their Graphs - Online Tutor, Practice Problems & Exam Prep

### Relations and Functions

#### Video transcript

State the inputs and outputs of the following relation. Is it a function? {$\left(-3,5\right),\left(0,2\right),\left(3,5\right)$}

State the inputs and outputs of the following relation. Is it a function? {$\left(2,5\right),\left(0,2\right),\left(2,9\right)$}

### Relations & Functions Example 1

#### Video transcript

Let's see if we can solve this problem. In this problem, we have three different graphs, and we're asked to determine which of the graphs are functions. We can select all that apply. Now, whenever you want to figure out whether or not a graph is a function, you can use the vertical line test. If you can draw a vertical line somewhere on the graph that crosses through more than one point, it is not a function. However, if the vertical line you draw only crosses through one point at most anywhere you draw it, it is a function. So let's see if we can figure this out.

Now, for this first graph, we'll try the vertical line test. If I draw a vertical line here, we only cross through one point. If I try a vertical line over here, we only cross through one point. If I try a vertical line through the center or close to the center, we only cross through one point. And it turns out, even though this graph expands in both directions, anywhere you would draw a vertical line would only ever cross through one point. So we can see that this first graph, Graph A, is an example of a function.

But let's try Graph B over here, and I'll go ahead and draw a vertical line. Notice the vertical line we drew crosses through more than one point. Since we see that we can draw a vertical line somewhere that crosses through more than one point, that means this is not an example of a function.

But now let's try the vertical line test on this graph over here. Well, if I draw a vertical line there, we only cross through one point. If I draw a vertical line here, we only cross through one point. And anywhere I would draw this vertical line would only ever cross through one point. So we would say that Graph C is an example of a function.

So, to list all of the graphs that are functions, we would say Graph A and Graph C are functions. That's the answer to this problem.

### Verifying if Equations are Functions

#### Video transcript

Welcome back, everyone. In the last video, we talked about relations and functions, and we specifically focused on graphs. Now, recall in the previous video that we had this special tool called the vertical line test where we could see if a vertical line ever passed through more than one point anywhere we would draw it on the graph. If this ever happened, we would say that we failed the vertical line test, in which case we would not be dealing with a function. However, if this did not happen, then we were dealing with a function. We talked about all of this in the last video. Now in this video, we're going to be taking a look at how we can verify equations as functions and there is a test for this, but the steps are a little more complicated. So pay close attention because I am going to walk you through them.

The first thing you want to do when dealing with an equation is always solve for y. This should be your first step when you have an equation. So if you have an equation and you've solved for y, what you then want to do is see if any x's will result in multiple y's. If this does happen, then you are not dealing with a function. However, if this does not happen, then you are dealing with a function. Now, notice for both of these tests, whenever we get an answer of yes, that means it's not a function. But whenever we get an answer of no, it is a function. So that's just something you can use to remember how these tests work. Now, let's actually put this test to use and solve an example.

So here we have these two equations, and we want to see whether or not either of these equations are functions. We'll start with equation a, which is this one on the left. We have y + 4 = 3x and our first step should always be to solve for y. So I'll take this 4 and subtract it on both sides of the equation to cancel the fours on the left. That will give me that y is equal to 3x - 4. And now that we've done this, we have solved for y. So the next thing I'm going to do is I'm going to try a bunch of x values to see if we have a situation where any of these x's result in multiple y's. So let's try an x value of 0. So we'll have that y is equal to 3 × 0 - 4. 3 × 0 is 0, and 0 - 4 is -4. Now let's try an x value of 1. We'll have y is equal to 3 × 1 - 4, 3 × 1 is 3, and 3 - 4 is -1. Now let's try a negative number like -1. So, we'll have y is equal to 3 × -1 - 4, 3 × -1 is -3, and -3 - 4 is -7. Now let's try an x value of 2. We'll just try one more value here to see what happens. So, we have 3 × 2 - 4, 3 × 2 is 6, and 6 - 4 is 2. Now notice what happens here. For every x value that we replace this x with, we always get only 1 y value as an output. And because of this, we could say that this is an example of a function because we only get 1 y value for each x value. And you may notice that this is actually a familiar form that we have this equation. We have the form y = mx + b. This is the slope-intercept form of a line that we've talked about in previous videos. So whenever you are dealing with a line that has a defined slope, and in this case, the slope would be positive 3, then you're always going to be dealing with a function in this situation. So that's just something you can remember when dealing with these types of problems. But now let's take a look at this equation on the right side. So for equation b, we have x² + y² = 25. And just like before, I'm going to solve for y. So I'll subtract x² on both sides of the equation and that'll get the x²s to cancel on the left, giving me that y² is equal to 25 - x². Now my next step here is going to be to take the square root on both sides of the equation, giving me that y is equal to ±√(25 - x²). So we've now solved for y, so this is the equation that we have. Now our next step is going to be to test out a bunch of different x values. So just like before, I'll start with an x value of 0. So we'll get that y is equal to ±√(25 - 0²). 0² is just 0, so all we're going to end up with is ±√25. And recall the square root of 25 is just 5. So because of this ± sign, we're going to end up with positive 5 and negative 5. But this is interesting. Notice how one x value gave us multiple y values. And we said whenever this happens that we are not dealing with a function. So, in this example, we can see that this is not a function because we already have an x value that gives us multiple y values. And something else you can remember with this equation is notice that we started with y². Whenever you have an equation where y has an even power that it's raised to, then it's always going to be not a function. So that's just something to keep in mind. Now neither one of these situations are scenarios that you're going to have to memorize, but it's just something that can help you if you want a quick shortcut to figuring out whether an equation is a function.

Now there's one more thing I want to mention before finishing this video and that is if you have an equation that is identified as a function, you can rewrite it using function notation. And when you use function notation, what you want to do is replace the y with f(x). So, for example, in this scenario that we had before where we had y = 3x - 4, you could take this y at the start and you could replace it with f(x) where x represents the inputs and f(x) represents the outputs. You could not, however, use this function notation in the other example because this was not identified as a function. So this is how you can verify equations as functions. Let me know if you have any questions and let's move on.

Is the equation $y=-2x+10$ a function? If so, rewrite it in function notation and evaluate at $f\left(3\right)$.

$f\left(3\right)=4$ , **Is A Function**

$f\left(3\right)=3$, **Is A Function**

$f\left(3\right)=1$, **Is A Function**

**Is NOT A Function**

Is the equation $y^2+2x=10$ a function? If so, rewrite it in function notation and evaluate at $f\left(-1\right)$.

$f\left(-1\right)=\sqrt{12}$, **Is A Function**

$f\left(-1\right)=12$, **Is A Function**

$f\left(-1\right)=\frac92$, **Is A Function**

**Is NOT A Function**

### Finding the Domain and Range of a Graph

#### Video transcript

Hi, everyone, and welcome back. Up to this point, we spent a lot of time talking about functions, and in this video, we're going to be taking a look at how we can find the domain and range of a function based on its graph. Now, when encountering finding the domain and range of functions, it can often seem a bit tricky. But in this video, we're going to be going over some analogies and examples that will hopefully make this topic a lot clearer. So let's get into this. When finding the domain of a graph, you're looking for the allowed x-values that you can have. And when finding the range, you're looking for the allowed y-values. Now, there's a little trick that you can use and that is known as the squish strategy. What you can do if you have the domain of a graph is you can take your graph and squish it down to the x-axis. So if you squish things down to the x-axis, that will tell you the domain. So let's say we have this graph down here. And notice how we took this curve, and we literally squished it down to the x-axis so only some sort of line or set of lines remains. This tells you the domain. Now when finding the range, you can take your graph and squish it down to the y-axis. So notice for our graph, we've squished it down to the y, and this tells you the range. Now, there are a couple of different ways we can write the domain and range of a graph. One of the common things is to use interval notation. Interval notation represents the domain and range using these brackets and parentheses. And you can actually see this in the example we have above. So notice that it looks like our domain goes from a value of negative 4 to positive 5 on the x-axis. So this is what we said the domain was. And for our range, it looks like we go from about negative 1 to positive 2 on the y-axis. So that's using interval notation. Now, there's another notation you can use, called set builder notation, and set builder notation uses these inequality symbols as opposed to the brackets and parentheses. Now, something you need to keep in mind whenever using either one of these notations is that whenever you see these brackets or inequality symbols with the bar underneath, that means that you want to include whatever value you're looking at. So let's say we have this graph for example. The values that you want to include are either closed dots or solid lines or curves. These are values that you always want to include in your domain and range. Now whenever you see these parentheses or inequality symbols without the bar underneath, this means that you do not want to include those values. And values that you do not want to include are situations where you see a hole in your graph. So anytime you see this hole or this open circle, that means you do not want to include it whereas anytime you see a solid dot or a solid line or curve, you do want to include it. So given all this information, let's see if we can solve an example. So here we're asked to determine the domain and range of the graph below and to express our answer using interval notation. Now what I'm first going to do is see if I can find the domain of this graph. And by the way, this is the same graph as you can see for both these diagrams. But for finding the domain, recall that what we want to do is take our graph and squish it down to the x-axis. So going down here, if we take this graph and we imagine squishing every single point down to the x-axis, we're going to get a graph that looks something like this. We'll have a line that goes from negative 4 all the way to an x value of 0 and then we'll have another line that goes from positive 1 all the way to 4. Now notice how there's an open circle here, so we have to include the open circle there as well. And then everywhere else, we have closed circles because these circles were all closed. So looking at this squished graph, our domain is going to go from negative 4 to 0 on our x-axis, and we need to include both these values since we have 2 closed circles. Now, I can see that another one of these lines is going to go from positive one because this is an x value of 1 right here to 4. So we're gonna have 1 to 4. We're going to include this positive one since we have a closed circle there, but our open circle is going to cause us to draw a parenthesis here because we do not want to include that value. Now notice how we have multiple intervals here. Whenever you have multiple intervals or a jump in your graph, you need to use something called a union symbol, which looks like this. So since we have multiple intervals, we need to put a union symbol between these to join the two intervals together. So this right here is the domain of our graph. Now let's see how we can find the range and recall when finding the range, we need to take our graph and squish it down to the y. So if I take this other graph we have drawn over here and I squish all these values down to the y, what we're first going to end up with is a line that goes from 1 to 3 on the y-axis and then we're going to end up with another line that goes from negative 3 all the way to negative 1, and we'll have an open circle at negative one since we have the open circle right there. So our range is going to go from negative 3 to negative one, and keep in mind we'll include the negative 3, but we do not include the negative one since we have an open circle, and our range will also go from positive one to 3. And we need to include both of these values since we have closed dots, or I should say we have a solid line or curve, but still, you want to include that value. And we'll put the union symbol, and then this is the range of our graph. So that is how you can find the domain and range of a function based on its graph. Let me know if you have any questions, and thanks for watching.

Find the domain and range of the following graph (write your answer using interval notation).

**Dom: $\left[-5,5\right]$ , Ran: **$\left[-4.4\right]$

**Dom: $\left[-2,2\right]$ , Ran: **$\left[-3,3\right]$

**Dom: $\left[-4,4\right]$ , Ran: **$\left[-5,5\right]$

**Dom: $\left(-5,5\right)$ , Ran: **$\left(-4,4\right)$

### Finding the Domain of an Equation

#### Video transcript

Hi, everyone. Welcome back. In the last video, we talked about how to find the domain and range of a function based on this graph. And in this video, we're going to be looking at how you can find the domain of an equation. Now, when finding the domain of an equation, it is something that is a bit more complicated than with a graph, but we are going to go over some of the common cases that you'll see that will hopefully make problems a lot more clear when you encounter them. So let's get into this.

Now recall, when finding the domain of a function, you're looking for the allowed x values that you can have, but there are going to be certain situations where you'll have restrictions on these x values. So when finding the domain of an equation, you need to first identify the values that will break the function. Any values that break your function are going to be restrictions that you'll have, and you need to be able to recognize these. And there are two common situations where we'll have restrictions. One of the common ones is whenever you have an x inside of a square root.

So the domain for x values inside a square root is going to be anything that does not make the inside of the square root negative. So you do not want to see a negative number underneath a square root. Let's take a look at this example. We're asked to find the domain of the function f ( x ) = x without graphing and to express our answer using interval notation. Now remember, you do not want the inside of the square root to be negative. So what that means is that our x values cannot be below 0. So any x values that are below 0 will make this a negative. Right? We can have 1 = 1. We can even have 0 = 0. But if we have - 1 , well, you can try plugging that into your calculator, you're going to get an error. This is not something you're allowed to have. So your x values cannot be below 0; therefore, your domain is going to be from 0 all the way to positive infinity. Your domain can equal 0, that's perfectly fine, but it just cannot be below 0. So this would be the domain of the function, and this is the answer for this example.

Now, if we actually take a look at the graph of the square root of x, notice that for the graph, this actually makes sense because notice that all of these values where x is above 0, these are all defined. But as soon as we look at the values where x is below 0, these values are not included. So, it makes sense that our graph would be from 0 all the way to positive infinity.

Now remember, this is one of the two common situations that you'll see where the domain has restrictions. The other common scenario is whenever you have x in the denominator of a fraction because you need to make sure that for the domain that your x values do not make the denominator equal to 0. You cannot divide by 0 in a fraction, so this is another situation you'll commonly run into. So let's take a look at this example. This example says, given the function f ( x ) = 2 x - 5 , find the domain using interval notation.

Now to find the domain of this function, the denominator cannot be equal to 0. So I'm just going to write out that x minus 5 cannot equal 0. And I can just solve this like a mini equation. So we'll take 5 and we'll add it to both sides, that'll get the fives to cancel, giving me x cannot be equal to 5. So that means the x values that make the denominator 0 are 5. So our domain cannot equal 5. So that means that our domain is going to be every value from negative infinity all the way up to 5, but not including the 5 because we can't actually equal this number. Then we're also going to have an interval that goes from 5 all the way to positive infinity. So, this basically says that we can have any real number except for the number 5, and that is the domain of the function.

So that is how you can find the domain of a function if you're given an equation rather than a graph. Hopefully, you found this helpful, and let me know if you have any questions.

Find the domain of $f\left(x\right)=\sqrt{x+4}$ . Express your answer using interval notation.

**Dom: **$\left[4,\infty\right)$

**Dom: **$\left[-4,2\right]$

**Dom: **$\left[2,4\right]$

**Dom: **$\left[-4,\infty\right)$

Find the domain of $f\left(x\right)=\frac{1}{x^2-5x+6}$ . Express your answer using interval notation.

**Dom: **$\left(-\infty,2\right)\cup\left(2,\infty\right)$

**Dom: **$\left(-\infty,\infty\right)$

**Dom: **$\left(-2,2\right)\cup\left(2,3\right)\cup\left(3,\infty\right)$

**Dom: **$\left(-\infty,2\right)\cup\left(2,3\right)\cup\left(3,\infty\right)$

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### Your College Algebra tutors

- In Exercises 1–30, find the domain of each function. f(x)=3(x-4)
- Without using paper and pencil, evaluate each expression given the following functions. ƒ(x)=x+1 and g(x)=x^2 ...
- Without using paper and pencil, evaluate each expression given the following functions. ƒ(x)=x+1 and g(x)=x^2 ...
- In Exercises 1–30, find the domain of each function. g(x) = 3/(x^2-2x-15)
- In Exercises 1–30, find the domain of each function. f(x) = 1/(x+7) + 3/(x-9)
- Let ƒ(x)=x^2+3 and g(x)=-2x+6. Find each of the following. See Example 1. (ƒ+g)(-5)
- Let ƒ(x)=x^2+3 and g(x)=-2x+6. Find each of the following. See Example 1. (ƒ-g)(4)
- In Exercises 1–30, find the domain of each function. f(x) = √(x - 3)
- Let ƒ(x)=x^2+3 and g(x)=-2x+6. Find each of the following. See Example 1. (ƒ/g)(5)
- In Exercises 1–30, find the domain of each function. f(x) = 1/√(x - 3)
- For the pair of functions defined, find (ƒ-g)(x).Give the domain of each. See Example 2. ƒ(x)=3x+4, g(x)=2x-6
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- In Exercises 1–30, find the domain of each function. g(x) = √(x −2)/(x-5)
- In Exercises 1–30, find the domain of each function. f(x) = (2x+7)/(x^3 - 5x^2 - 4x+20)
- In Exercises 31–50, find f−g and determine the domain for each function. f(x) = 2x + 3, g(x) = x − 1
- In Exercises 31–50, find f/g and determine the domain for each function. f(x) = x -5, g(x) = 3x²
- Use the graph to evaluate each expression. See Example 3(a). (ƒ-g)(1)
- In Exercises 31–50, find fg and determine the domain for each function. f(x) = x -5, g(x) = 3x²
- Use the graph to evaluate each expression. See Example 3(a). (ƒg)(1)
- In Exercises 31–50, find fg and determine the domain for each function. f(x) = 3 − x², g(x) = x² + 2x − 17
- In Exercises 31–50, find f−g and determine the domain for each function. f(x) = 3 − x², g(x) = x² + 2x − 16
- In Exercises 31–50, find ƒ+g and determine the domain for each function. f(x) = 3 − x², g(x) = x² + 2x − 15
- In Exercises 31–50, find ƒ+g and determine the domain for each function. f(x) = √x, g(x) = x − 4
- In Exercises 31–50, find f/g and determine the domain for each function. f(x) = √x, g(x) = x − 4
- In Exercises 31–50, find ƒ-g and determine the domain for each function. f(x) = 2 + 1/x, g(x) = 1/x
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- In Exercises 31–50, find f/g and determine the domain for each function. f(x)= = 8x/(x - 2), g(x) = 6/(x+3)
- In Exercises 31–50, find f−g and determine the domain for each function. f(x) = √(x +4), g(x) = √(x − 1)
- In Exercises 31–50, find ƒ+g and determine the domain for each function. f(x) = √(x +4), g(x) = √(x − 1)
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- For each function, find (a)ƒ(x+h), (b)ƒ(x+h)-ƒ(x), and (c)[ƒ(x+h)-ƒ(x)]/h.See Example 4. ƒ(x)=-x^2
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- In Exercises 51–66, find a. (fog) (2) b. (go f) (2) f(x)=4x-3, g(x) = 5x² - 2
- Let ƒ(x)=2x-3 and g(x)=-x+3. Find each function value. See Example 5. (ƒ∘g)(4)
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- In Exercises 59-64, let f(x) = 2x - 5 g(x) = 4x - 1 h(x) = x² + x + 2. Evaluate the indicated function withou...
- In Exercises 67-74, find a. (fog) (x) b. the domain of f o g. f(x) = 2/(x+3), g(x) = 1/x
- In Exercises 67-74, find a. (fog) (x) b. the domain of f o g. f(x) = x/(x+1), g(x) = 4/x
- Given functions f and g, find (a)(ƒ∘g)(x) and its domain. See Examples 6 and 7. ƒ(x)=-6x+9, g(x)=5x+7
- Given functions f and g, find (b)(g∘ƒ)(x) and its domain. See Examples 6 and 7. ƒ(x)=8x+12, g(x)=3x-1
- Given functions f and g, find (b)(g∘ƒ)(x) and its domain. See Examples 6 and 7. ƒ(x)=√x, g(x)=x+3
- In Exercises 75-82, express the given function h as a composition of two functions ƒ and g so that h(x) = (fog...
- In Exercises 76–81, find the domain of each function. g(x) = 4/(x - 7)
- Given functions f and g, find (a)(ƒ∘g)(x) and its domain. See Examples 6 and 7. ƒ(x)=x+2, g(x)=x^4+x^2-4
- In Exercises 75-82, express the given function h as a composition of two functions ƒ and g so that h(x) = (fog...
- Given functions f and g, find (b)(g∘ƒ)(x) and its domain. See Examples 6 and 7. ƒ(x)=2/x, g(x)=x+1
- Given functions f and g, find (a)(ƒ∘g)(x) and its domain. See Examples 6 and 7. ƒ(x)=2/x, g(x)=x+1
- Given functions f and g, find (b)(g∘ƒ)(x) and its domain. See Examples 6 and 7. ƒ(x)=√x, g(x)=1/(x+5)
- Use the graphs of f and g to solve Exercises 83–90. Find (fg) (2).
- Use the graphs of f and g to solve Exercises 83–90. Find(g/f)(3)
- In Exercises 89–90, express the given function h as a composition of two functions f and g so that h(x) = (f ○...
- Use the graphs of f and g to solve Exercises 83–90. Graph f-g.
- In Exercises 91–94, use the graphs of f and g to evaluate each composite function. (fog) (-1)
- Let ƒ(x) = 3x^2 - 4 and g(x) = x^2 - 3x -4. Find each of the following. (f+g)(2k)
- Let ƒ(x) = √(x-2) and g(x) = x^2. Find each of the following, if possible. (ƒ ○ g)(x)
- Let ƒ(x) = √(x-2) and g(x) = x^2. Find each of the following, if possible. (f ○ g)(-6)
- Use the table to evaluate each expression, if possible. (f-g)(3)
- The graphs of two functions ƒ and g are shown in the figures. Find (g∘ƒ)(3).