The following equations implicitly define one or more function...

Question

The following equations implicitly define one or more functions.

c. Use the functions found in part (b) to graph the given equation.

x+y³−xy=1 (Hint: Rewrite as y³−1=xy−x and then factor both sides.)

Accepted Answer

Hi everyone, let's take a look at this practice problem. This problem says to graph the functions Y is equal to F1 of X, Y is equal to F2 of X, etc. that are implicitly defined by the equation 4 X plus Y minus 2x multiplied by Y is equal to 8. We need to rearrange the terms of the equation as follows, and then factor both sides. And we're given the expression Y cubed minus 8 is equal to 2 x multiplied by Y minus 4 X. Now, the first part of this question wants us to take our equation, which is 4 X. Plus y cubed minus. 2 X multiplied by Y is equal to 8, and rearrange it so that we get Y minus 8 is equal to 2 X multiplied by Y minus 4 X. So to do this, the first thing we'll do is add 2 X multiplied by Y to both sides of the equation. Then we'll subtract 4 X from both sides of the equation. And then we'll subtract 8 from both sides of the equation. In doing so, we will get Y cubed minus 8 is equal to. 2 X multiplied by Y minus or X. Now, the next step is to factor both sides of the equation. Now, on the left hand side, we have the difference of two perfect cubes. And so recall that when you factor that type of expression, you're going to get Y minus 2, and that quantity is going to be multiplied by the quantity of Y2d. Plus 2 Y + 4. In quantity. And then this is going to be equal to. On the right-hand side, I can factor out a 2 X out of both terms, that will have 2 X multiplied by the quantity of Y minus 2 in quantity. Now, to solve this equation, we noticed that on both sides of the equation, we have a factor of Y minus 2. So if Y minus 2 is equal to 0, then this statement would be true. So one of the solutions will be that Y minus 2 is going to be equal to 0. So that means that Y is going to be equal to 2. That means our first function, which we'll call F1 of X. It's simply going to be equal to 2. So now the other possibility is that Y is not equal to 2. Then what remains must be true, since we can cancel out a Y minus 2 from both sides. And so we will have Y2 + 2 Y + 4 is equal to 2 X. Now, we can subtract 2 X from both sides of this equation, and we will get Y2 + 2 Y + 4 minus 2 X is equal to 0. And so now we can solve this equation using the quadratic formula. So we call for the quadrag formula, we're going to have Y. is equal to -2 plus or minus the square root of the quantity of 2 squad. -4, multiplied by 1, multiplied by the quantity of 4 minus 2 X in quantity. And that whole quantity is divided by the quantity of 2 multiplied by 1. Now, underneath the square root sign, we can factor out a 4 from both of the terms, and then we'll have the square root of 4, which is 2, and so we'll be able to cancel out the 2 in the denominator, since we'll have a factor of 2 in both terms in the numerator. So that means that this is going to be equal to -1 plus or minus the square root. Of the quantity of 1 minus the quantity of 4 minus 2 X. So some find the quantity inside the radical. This is going to be equal to -1 plus or minus the square root of the quantity of 2 X minus 3. So this means we have two solutions here, and so one solution will be Y is equal to -1 plus the square root of the quantity of 2 X minus 3 in quantity, and the other solution will be Y is equal to minus 1 minus the square root of the quantity of 2 X minus 3. So that means that we will have F2 of X. To be equal to minus 1, plus the square root of the quantity of 2 X minus 3 in quantity, and we'll have F3 of X. To be equal to -1 minus the square root of the quantity of 2 X minus 3 in quantity. So now that we have all of our functions for our Fs defined, we can actually plot all of them on our graph. So, here we're going to be using our graphing utility. And when we use our graphing utility, we get the following graph. Which has a horizontal line. At Y is equal to 2. And then we have a parabolic curve that is opening up towards the positive X-axis, where the top half of that is Due to F2, and the bottom half is due to F3. So, this is the graph that we were actually looking for for this problem. So I hope that this explanation has been useful, and I'll see you in the next video.

The following equations implicitly define one or more functions.
c. Use the functions found in part (b) to graph the given equation.
x+y³−xy=1 (Hint: Rewrite as y³−1=xy−x and then factor both sides.)

Key Concepts

Implicit Functions

Factoring

Graphing Functions

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