In Exercises 73–74, use the graph of the rational function to sol...

Question

In Exercises 73–74, use the graph of the rational function to solve each inequality.

Graph of a rational function showing asymptotes and a point at (-9/5, 0).

1/4(x + 2) ≤ - 3/4(x - 2)

Graph of the function f(x) = (x + 1)/(x^2 - 4) with labeled axes.

Accepted Answer

Everyone welcome back in this problem. Using the following graph, we're asked to solve the given inequality. The inequality is 1/5 X plus three is less than or equal to negative 4/5 X minus three. So let's just write out our inequality 1/5 X plus three and just pay attention to the brackets. Okay. The X plus three is in the denominator and same with the X minus three on the other side. I need a 4/5 x minus three. Alright, so we're asked to use the graph to solve this inequality. If we look at the graph, Okay the function in the graph we're given is five X plus 9/5 X squared minus nine. Okay so what we wanna do is we want to try to manipulate our inequality so that's related to this function then we'll be able to use the graph to solve the inequality. Alright so let's start by moving everything to the left hand side. So we get 1/5 X plus three plus four Over five X -3. And this is going to be less than or equal to zero. So when we're working with inequalities we can add and subtract from both sides. Just like a regular equation. Okay, we keep the inequality the same. The only time that we need to worry is if we're multiplying or dividing by a negative, okay then the sign is going to switch but otherwise when we're just adding and subtracting, we do it just like an equation. Alright so we've moved that over Now, we want to get it to look like this. So we want kind of one term instead of having two separate terms. So let's try to find a common denominator. So on the left term we're gonna multiply by X -3. So we multiplied by x minus three and we have five X plus three X minus three in the denominator. Okay so we've multiplied top and bottom by x minus three. So we haven't changed the function and same thing for the second term. We're gonna multiply by X plus three and this is gonna give us a common denominator we're gonna have five X plus three X minus three for both terms. And this is all still less than or equal to zero. Alright, now that we have this common denominator, we can add them together. So in the numerator we get X -3 plus four X plus three divided by five X plus three X. In history. And again we're trying to get to this function that's represented in the graph let's go ahead and simplify and expand in the new murder. Okay so we get X -3 plus four X plus 12. And in the denominator if we multiply, Okay we have X plus three times X minus three, you'll notice that this is a difference of squares and this gives us X squared minus nine which is the denominator we have in the function we're trying to get to so that's awesome. Let's just simplify one more step for that numerator. We have X plus five X. Or sorry X plus four X. Which is gonna give us five X. And then we have negative three plus 12. This is going to give us plus nine. And then we have divided by five X squared minus nine is less than or equal to zero. And now this function on the left hand side is the function of the graph. Hm. So we want this function that is graft to be less than or equal to zero. So we want to find X values where the graph is on or below the X axis. Okay. If the graph is on the X axis that means that that function equal zero. Okay. So we're satisfying this equal part of our less than are equal. And when it's below the X axis that means that that function those y values are negative. There are less than zero. So we're gonna satisfy that less than part. Okay. So we want to take a look at our function our graph of our function and figure out all of the X values where the graph is on or below the X axis. All right, let's do it. So here's our function. Here's our graph. Where is this thing below the X axis or on the X axis we can see on the left hand side here. This part that I'm tracing of our graph. This is all below the X axis. These are negative values. And we also have here we have this point negative ninth is zero where we're on the X axis and then it continues downward below the X axis. All right. So if we're looking for the corresponding X values where this is happening, you see that We have X values to the left of this point here, which is -3. So that's gonna be from negative infinity to negative three. Now at negative three, this function kind of goes off to negative infinity and jumps back up to positive infinity. And so it doesn't look like the functions to find that negative three. So we're gonna use a round bracket to indicate that we're not including negative three because we don't know what happens right at negative three. And then we want to include our other interval as well. Now are other interval. Here we go from here, 9/5 All the way over to positive three. Now we're gonna use a square bracket on the negative 9/ because we have the point negative 9 50 0. Okay, We want to know where our function is. Less than or equal to zero. Okay, It's equal to zero at negative nine fist. So square bracket to indicate that we include that point for that value. Okay. And this goes all the way over to positive three. We're gonna use a round bracket on positive three. Okay. For the same reason we did with negative three and so there we have it. Can we have our solution the X. Values that solve this inequality are gonna be negative infinity to negative three. And the union of negative nine fists to three where we include negative 9/5. So we look and we have answer choice. D. Thanks everyone for watching. I hope this video helped see you in the next one.

In Exercises 73–74, use the graph of the rational function to solve each inequality.
1/4(x + 2) ≤ - 3/4(x - 2)

Key Concepts

Rational Functions

Asymptotes

Inequalities

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