 In this video, we provide the solution to question number eight from practice exam number two for math 1210, and we're asked to find the largest number delta possible such that when the absolute value of x plus one is between zero and delta, that'll guarantee that the absolute value of three minus four x minus seven is less than one fourth. So this is an epsilon delta type question. Let's make sure we can label all the appropriate parts right here. So this is our function in play right here three minus three minus four x that is our function f of x it's our y coordinate there. This number seven is the limit, which we call L in this one fourth there's our epsilon deltas will we have to find this one right here we might want to rewrite x plus one is x minus a negative one for which we then identify that a is our value of negative one. So we want to know in a symmetric manner how much to the left or right of negative one should we be to guarantee that our function is within one fourth away from seven right here. So if we want to solve this algebraically that would be very appropriate we have to solve we have to solve the equation f of x equals L plus or minus epsilon there's two equations right there so one of them we three minus four x is equal to seven plus instead of one fourth I'm going to put as a decimal point two five like so I mean you can leave it as a fraction if you want to whichever you prefer it doesn't make much of a difference honestly, as the answers are given as fractions it's probably best to keep it as a fraction. So we're going to three minus four x is equal to well with seven we might have to write as 28 fourths. And so we end up with 29 fourths right here. Let's subtract three from both sides, we're going to get negative four x is equal to if we take, if we take three, that can be written as 12 fourths. And so that would then become subtracting 12 fourths from 29 fourths. We can write that there minus three here minus 12 fourths right there. That will end up with a 17 force, in which case then if we divide both sides by negative four, we end up with x is equal to negative 17 over 16. So that's one of the values we need to keep track of this is our so called a one. We have to also do the other equation three minus four x is equal to seven minus a fourth, which that's going to become 27 force, we still are going to subtract three from both sides, which will then translate to become a 12 fourths right there. This gives us that negative four x is then equal to subtracting the 12, we're going to get 15 force, for which then when we divide here by negative four again we're going to get x equals negative 1516, which is our a two. And so then delta, as you'll recall delta is going to be the minimum. Whoops. It's the minimum of the absolute value of a minus a one with the absolute value of a minus a two. Like so, for which if we subtract, if we subtract a one from our a value which is negative one, you're going to end up with a 116. Like so, if you do a two minus a one or minus a excuse me, you're going to get also 116. Okay, we're taking absolute values, this thing's going to be positive. And so they're actually both the same and so we end up with 116. And that's going to be our delta value, which then makes the correct answer be now that's sort of like the long drawn out way you can do this one that's going to work for general functions that you could see here. Now some variants I want to mention here is that the function itself might actually be given to you in graphical form. And which case that basically tells us that the a one and a two are probably already given to you that simplifies a lot of this process because we don't have to solve the equation that one. So if you're given a graph, you know, scream hallelujah because that makes this question a whole lot easier for us. Another thing I want to mention to us is that since our function here is linear, there's a nice little shortcut here that if you ever given. If you're given epsilon, and you want to find delta delta is just going to equal epsilon divided by the absolute value of the slope. That is to say in this one we have one fourth divided by the slope of the line which is negative one fourth there. So we end up with 116. This is a character this is characteristic of lines that to find the delta for a line you just take epsilon and divided by the slope taking absolute value that only works for linear functions like y equals three minus four x but that's something we could have applied in this one to dramatically simplify the process.