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Use the graph of the function to answer the questions. Answers may vary.a. Estimate $f(1)$.b. Estimate $x$ such that $f(x)=5$.c. State the domain and range for $f(x...

Question

Use the graph of the function to answer the questions. Answers may vary.a. Estimate $f(1)$.b. Estimate $x$ such that $f(x)=5$.c. State the domain and range for $f(x)$.d. Estimate the vertical and horizontal intercepts for $f(x)$

Use the graph of the function to answer the questions. Answers may vary. a. Estimate $f(1)$. b. Estimate $x$ such that $f(x)=5$. c. State the domain and range for $f(x)$. d. Estimate the vertical and horizontal intercepts for $f(x)$



Answers

The graph of a function $f$ is given. (a) State the value of $f(1)$ (b) Estimate the value of $f(-1)$ (c) For what values of $x$ is $f(x)=1 ?$ (d) Estimate the value of $x$ such that $f(x)=0$ (e) State the domain and range of $f$. (f) On what interval is $f$ increasing?

All right, toe. Let's go ahead and answer this question. So the question is we have Quint IQ X to the fifth minus X cubed plus two. And then it is limited between the region. Negative 1 to 1. The question is, graph it and approximate the maximum and the minimum value. And the other part is use calculus and what you've learned to precisely locate the exact value off the maximum and the minimum so as usual, because the region is limited, we need to keep in mind the extreme value theory, Um, saying that if a maximum or minimum occurs, we have to check the minimum and the maximum value, not the minimum and maximum. Excuse me, the endpoints, negative one and one. So we're going to check what f of one in efforts Negative one are going to be okay. If you plug in these numbers, you can see that the negative one minus negative one cancels each other out. So you will get a two. And the same thing happens when you plug in that one. Okay, one. My next one is zero so f of 1.5 of negative one are both to Okay, so it's neither a maximum or minimum right away, because we know that this is not a straight line. Okay, so we're going to now take a look at the critical points. If the derivative is equal to zero, you know that those points are likely to be a maximum and the minimum. And this is a polynomial with, um, odd power. So we know that we're not gonna be dealing with a situation where, um the maximum and the minimum bounces off the the X axis and such. So it's highly likely, but it's not a saddle, but we gotta check it anyways. Okay. So, first, let's take a look at the graph I already grafted for you. X cubed minus x x to the fifth minus X cubed. Plus two is given right there. And to make it look a little bit better, X equals two negative. 1 to 1. I do this negative one positive one. I know that it is this region to that region that I'm focusing on. So the maximum value happen some around here. Let's see if I can help. I can. Maybe around there. So it seems like 2.2186 or so is going to be my maximum. The minimum here is 1.814 So let's see if we can get those two numbers. And as you can see, it's indicating that it's in metric. So possibly the solution is going to be a very, um, plus or minus something kind of number where if we plugged it in, we get these heights. Okay, so that's one thing that we could observe. All right, so let's go ahead and try to do this, um, analytically. So we're going to find the derivative first. This one shouldn't be that big of a deal by now. Have prime off X is five x to the fourth minus three X square to just become zero. We're gonna let this equal to zero. So X equals to zero is, of course, going to be a saddle. As we saw over here, it decreases but decreases again. So it's not a critical point. So it actually did show up my dad before the explanation that I made, um so I factored out on X squared and did that calculation. What's the other part? It is going to be plus ra minus the square root off 3/5. So those are the critical points. If you take a look, you can see that it's going to give you a maximum here, minimum there. If we started off with the analytic process, I could have probably find the second found the second derivative. Plug these numbers to see if it's going to be positive or negative, because if you get a positive, sick and derivative, we know that we will end up having a minimum because it's can't give up and vice versa. If it's can't keep down, it's a maximum. Okay, moving on. So we don't want to end here. We want to actually find the maximum and minimum, not where it's located at. So we're going to plug in these two numbers into F. So I am only going to do the case where it's the square root of 3/5. Just to show you the trick off what we can do to make the process a little bit simpler. Okay, so if we plug this number in, we can do the calculation, of course, but to make it simpler, you can see that I can factor out on X cubed from both of these expressions. So X cubed times X squared, minus one plus two. Why did I do that? Well, X squared is a nice number. It's 3/5. Right? So this expression right here inside the parentheses, it's 3/5 minus one, which is negative to fifth X. Cubed is also not that difficult to calculate because I know that X squared is to fifth. I meant 3/5. So this is 3/5 square root of 35 So if I go through the calculation, it is going to simplify to negative 6/25 the square root of 3/5. I am not going to rationalized the denominator, because to me, this honestly looks simple enough. Plus two. And this is the expression that you will get over here. Okay. If you plug in f off negative, you, uh, the skirt off 3/5. It has to go all the way there. You will get the positive version of this expression plus two. Okay. And then these air going to be the precise location off the maximum and the minimum, and then you can see that if you plug it in a calculator. It will actually be these two numbers that you see right there. Okay. And this is how you solve this problem?

All right. So the problem we're gonna be working on is, uh, when we were given a function, and we have to answer questions as it relates to, um, the given function. One thing you gotta remember is that when we have a function, we always have to remember that all functions are, uh, written is f of X equals. Why? The reason this is important is because we have our inputs, which is our X values. And then we have our outputs, which is R y values where we put in, we get out. So if we go back to this particular question says state the value of F one. So what I'm thinking when I'm doing a problem like this is my input is X equals one. So what must I do? Well, since I don't have an equation, I'm going to I just have a graph. What we're gonna do is go Toe X. I look at X equals one, go up to my function over to why, and I get, um, why is equal to three. So for that first problem, we have number one uh, f at one gives us the value of three for the second problem listed there. We have estimate the value of effort. Negative one. All right. So, again, there's my input. X equals negative one. I go over the negative one of my X axis, which is right here, and I have to estimate it because it doesn't land right on one of the tick marks. So what we're gonna need to do is just kind of estimate it's a little bit below, um, negative 0.5, as you can see there. So I'm gonna say that's maybe negative 0.33 or negative 1/3. So when we input a negative one, we get negative. 1/3 out. So for number two, I must say f of negative one is going to give us approximately negative 0.33 You could also wrote that is 1/3 if you'd like number three. Number three. We have for what values of X is f of X equal one. So now we have to work through this back words Clearly, we don't have our input value, but we do have an output value, as you can see, uh, they're with one. So what we're gonna do is we're gonna start backwards. We're start at one, and at one at this function is gonna give us the X value there. Also, If I were to stick a zero back into this equation are back into this function I would get one is my output. So for letter C, we would have f of X at one is gonna give us an X value of zero. Okay. Ah, the fourth question there. Fourth question access to estimate the value of X, such that f of X is equal zero again. We're gonna look where y is equal to zero. So, uh, my wife values equal to zero when my function is on the X axis. So this is why equal zero. You can see my function lies or crosses the X axis right here, which is a little bit more than halfway again. I'm going to say that's about 1/3. So we'll say negative 0.67 So f of X, where y is equal to zero. I'm gonna say X is approximately, uh, negative 0.6 seven. For the fifth question we have there fifth question asked us to state the domain and range of our domain values are all of our possible inputs that make this function true. Arrange values, air all the possible outputs we get. So if we look here, the things that make this function true are negative to all the way to, uh, looks like positive for the things that we get out of our function. When we enter any of those values go from negative one all the way up to a positive three. Okay, so the way we would write that is we would write our domain. I'm gonna say our domain is so our domain is all of the possible X values from negative. Want a negative too, All the way to a positive four. So a negative too. Too positive. Four. Our range values air all the possible outputs. So our range values would go from negative one all the way up to a positive three. Negative one all the way to positive three. The last question their states on what intervals at f increasing. The easiest way to see that is just to identify. Look at your X access. Identify where the function itself is increasing going up from left to right. So as I'm looking at my ex excess. I can see my function is going up as I continue along the X axis up until I get to the value of one. Then it starts to decrease our functions. Gonna increase from negative too, All the way to a positive one. So we're increasing from negative too, to a positive one.

Okay, so the question that we are going to be looking at today is pertaining to functions and how to interpret them on a graph. So, as you can see in the right hand corner of our textbook and in this video, we have the graph. And so we are going to be using this, um, graph to answer our questions. And particularly we're going to be looking at this red curve which represents our function. So let's begin in part A. We are looking at f of negative one. So what is this Syntex? Remember that the function is represented as F of X is equal to y where X isn't input. And why is the output? And so in this case, are X is negative one. So what is the output at negative one? So let's go on our graph and find her answer so we can go. Here's the X axis. Here's the Y axis. Okay, so the X is the horizontal. So we are going to start at zero, and we're going to go to the left one. Now, what is the function at negative one? That is me asking. Where does this pink or red curve intersect with this green line. And that is when Why is negative two. So that is our answer. Part B Part B is very similar, except we have ff two. So in this case, the access to our input. So what is our way? So let's do the same thing. Start at zero. You go 12 here. Now what is why one X is too so we can go up and down the screen line. And where does the red intersect that green that is here It is not not quite at three yet. And so we would call out about 2.8 Good part C. Part C is asking when f f X equals two. Then what is X equal to this is like the opposite. We are given ry value and we are looking for X. So essentially, when y is equal to two, where does this pink line? I'm sorry. Where does this red curb intersect? This blue dash line When? Why is he going to? Well, it enters us twice here and here. And so those are answers. X is now negative three and X equals one. Good. So let's go on to part D Part D is asking when Alphabet X equals zero. What is exe? Okay, so it's very similar to the previous question that we are looking at when why is equal to zero. So that is the X access. Essentially. Now, where does that bread curve intersect with this new blue line again intersex twice here and here. And so for this one, we will call that about negative 2.5. You can see it's somewhere in the middle between negative two and three and also here the second one. We'll call that about 0.3. It's not quite halfway. Okay, now to Part E. We are finding the domain and range. So what is the domain? We will just call the domain, do you to be short. So the domain is all the input variables within the X axis on the horizontal. So what values does the function take from the X axis? Essentially, we're going to be looking at the endpoints of our function here and here. That is our domain. It's kind of like the width of the function. Let's call this like the width. How far does are function extend to so it starts here and it stops here. So that is our domain. So we can see that as at negative. 12 negative. Three and 123 That is our domain. Now what is our arrangement? Call that are our range is, like, kind of like the opposite. We are looking at the height of our function. We see that the range essentially takes in the all the output variables of our function the y. So you're going to be looking up and down and you're going to be looking at the minimum of our function, which is here. That's the lowest point. Our function goes, and our maximum, which is here, that's the highest point of function goes. So where are those two that is here at negative two? Remember, we are not caring about this. We're talking about this. So that is here. Negative too. And then up here. 123 That is our range. Good. Now where is our function? Increasing. So how do we do that? Well, we can look at where the function changes direction when it is increasing. It is essentially going up when it's decreasing. It is going down so you can see here it is decreasing. Right. But around here, the minimum is where it changes. It's increasing. And so our function increases again. For this, we will look at the X axis, so increases from negative one, 212323 That is where our function increase in the sea from negative. One, 23 is where function increases. And you find that by looking at the minimum the lowest point, it starts to change. Decrease, increase. Mhm. Um, I hope you were able to solve this problem and learn a lot. So until next time.

So the question here gives us a graph of a function of F here, so F is just going to be any particular function here, which is shown in the graph, and it wants us to answer the following questions. So for a year, it says, What is the value of f of one? So f of one is essentially when X is equal to one, So this is going to be equal to three year for be here. We have wants us to estimate the value of F of negative one. So f of negative one is, of course, when X is equal to negative one. And what we get is that that's going to be approximately negative 0.25 for see here, it says, for what values of X is, um, f of X equal to one here. So essentially, this is the same thing as saying What is the values of X when y is equal to one? So what we get is that X equals zero and X equals three is when this particularly happens, Um, for D here it says, um, it wants us to estimate the value of X, such that F of X is equal to zero. So when is y equals zero? We get that this occurs at X equals negative 0.8, um, for e here it says the domain, um well wants us to state the domain and range of functions. So the domain, if we recall, is going to essentially be the value of X and the range is gonna be the values of why, in this particular graph, So the domain of F is gonna be between negative two and four. Um, the range of why of f rather is gonna be negative one and three here and lastly for F here, it basically wants us to find the intervals at which F is going to be increasing. So in the graph, it's gonna be the points where f is gonna be increasing, like such. So this starts increasing on well between the values of negative two and one here. After that, it's going to be, um, patrolling or decreasing this particular case. So this is gonna be the answer for this particular problem


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