+2446 You are wondering how to graph the following equation of y=√2x+2√x2−1−√x−1. First, we should determine the domain to see what the real inputs are:
| y=√2x+2√x2−1−√x−1 | Factor out a 2 from 2x+2 |
| y=√2(x+1)√x2−1−√x−1 | Also, x^2-1 can be factored into (x+1)(x-1). |
| y=√2(x+1)√(x+1)(x−1)−√x−1 | Split the 2 out of the radical. |
| y=√2√x+√(x+1)(x−1)−√x−1 | |
Let's think about which values for x will result in a nonreal answer. The square root of a negative number results in a nonreal answer. Let's figure out when the radicand is less than 0. Let's tackle the easy one first.
| x−1<0 | Add 1 to both sides of the equation. |
| x<1 | |
Therefore, we know that when x<1, it is not apart of the domain. Now, let's tackle the hard one:
| (x+1)(x−1)<0 | Let's calculate when both factors are less than 0 and then calculate solutions. | ||
| Add or subtract 1 from both sides. | ||
| |||
I have not solved the inequality completely, but we know that the solutions must be something less than 1. However, we have already determined that a number less than 0 would result in a non-real output, so there is no reason to consider this inequality:
Domain:{f(x)∈R|x≥1}
Now that we have determined the domain of the function, let's begin plotting some points. By determining the domain, we are eliminating the possibility of plugging in a number not in the domain. However, before we begin plotting, we should be familiar with the parent function of f(x)=√x. I have a picture below that illustrates it:
Let's plug in the first available value into the equation, 1. That is part of the domain, so let's plug it in and see what we get:
| f(x)=√2x+2√x2−1−√x−1 | Replace every instance of x with one. |
| f(1)=√2∗1+2√12−1−√1−1 | Simplify. |
| f(1)=√2+2√0−√0 | Of course, √0=0 |
| f(1)=√2 | |
Therefore, one point on the graph is (1,√2). Let's find another. Let's plug in 2:
| f(x)=√2x+2√x2−1−√x−1 | Replace every instance of x with two. |
| f(2)=√2∗2+2√22−1−√2−1 | Simplify |
| f(2)=√4+2√3−√1 | √1=1, of course. Now, let's simplify this monstrosity√4+2√3 |
| √4+2√3+√32−3 | Maybe you can that I am not changing the value of this expression by adding those 2 terms. The significance of which will become clear soon. Simplify. |
| √√32+2√3+1 | It might be difficult to notice this, but the radicand is actually a perfect-square trinomial. |
| √(√3+1)2 | The square root and the square function undo each other. |
| √3+1 | Now, let's replace √4+2√3 with its simplified value. |
| f(2)=√3+1−1 | |
| f(2)=√3 | |
Therefore, another point is (2,√3). Let's one more example. I know ahead of time that x=5 is a good point to plug in:
| f(x)=√2x+2√x2−1−√x−1 | Just like before, substitute x for 5. |
| f(5)=√2∗5+2√52−1−√5−1 | Simplify. |
| f(5)=√10+2√24−2 | Let's try to do the same process like above to simplify the radical: |
| √10+2√24 | |
| √10+2√24 | Rearrange the terms to put them in the form of a^2+2ab+b^2. Before we do that, however, we must simplify the square root of 24. |
| √10+4√6 | Now, do the same process as above |
| √10+4√6+√62−6 | Rearrange the terms and simplify. |
| √√62+4√6+4 | Yet again, this is a perfect square trinomial. Factor it. |
| √(√6+2)2 | Just like above, the square root and the square functions cancel. |
| √6+2 | Plug this in. |
| f(5)=√6+2−2 | |
| f(5)=√6 | |
Here's a third point (5,√6). Not every integer will be nice solutions. These, however, are examples of the nicest examples that you will get. After plotting these 3 points, connect them loosely like the shape of the parent function. I picked those points because I knew that they all would have a nice answer. Other points you plug in may not have as nice of a solution.
If you would like to reference a graph, I have one for you! Click here to view it:
+2446 You are wondering how to graph the following equation of y=√2x+2√x2−1−√x−1. First, we should determine the domain to see what the real inputs are:
| y=√2x+2√x2−1−√x−1 | Factor out a 2 from 2x+2 |
| y=√2(x+1)√x2−1−√x−1 | Also, x^2-1 can be factored into (x+1)(x-1). |
| y=√2(x+1)√(x+1)(x−1)−√x−1 | Split the 2 out of the radical. |
| y=√2√x+√(x+1)(x−1)−√x−1 | |
Let's think about which values for x will result in a nonreal answer. The square root of a negative number results in a nonreal answer. Let's figure out when the radicand is less than 0. Let's tackle the easy one first.
| x−1<0 | Add 1 to both sides of the equation. |
| x<1 | |
Therefore, we know that when x<1, it is not apart of the domain. Now, let's tackle the hard one:
| (x+1)(x−1)<0 | Let's calculate when both factors are less than 0 and then calculate solutions. | ||
| Add or subtract 1 from both sides. | ||
| |||
I have not solved the inequality completely, but we know that the solutions must be something less than 1. However, we have already determined that a number less than 0 would result in a non-real output, so there is no reason to consider this inequality:
Domain:{f(x)∈R|x≥1}
Now that we have determined the domain of the function, let's begin plotting some points. By determining the domain, we are eliminating the possibility of plugging in a number not in the domain. However, before we begin plotting, we should be familiar with the parent function of f(x)=√x. I have a picture below that illustrates it:
Let's plug in the first available value into the equation, 1. That is part of the domain, so let's plug it in and see what we get:
| f(x)=√2x+2√x2−1−√x−1 | Replace every instance of x with one. |
| f(1)=√2∗1+2√12−1−√1−1 | Simplify. |
| f(1)=√2+2√0−√0 | Of course, √0=0 |
| f(1)=√2 | |
Therefore, one point on the graph is (1,√2). Let's find another. Let's plug in 2:
| f(x)=√2x+2√x2−1−√x−1 | Replace every instance of x with two. |
| f(2)=√2∗2+2√22−1−√2−1 | Simplify |
| f(2)=√4+2√3−√1 | √1=1, of course. Now, let's simplify this monstrosity√4+2√3 |
| √4+2√3+√32−3 | Maybe you can that I am not changing the value of this expression by adding those 2 terms. The significance of which will become clear soon. Simplify. |
| √√32+2√3+1 | It might be difficult to notice this, but the radicand is actually a perfect-square trinomial. |
| √(√3+1)2 | The square root and the square function undo each other. |
| √3+1 | Now, let's replace √4+2√3 with its simplified value. |
| f(2)=√3+1−1 | |
| f(2)=√3 | |
Therefore, another point is (2,√3). Let's one more example. I know ahead of time that x=5 is a good point to plug in:
| f(x)=√2x+2√x2−1−√x−1 | Just like before, substitute x for 5. |
| f(5)=√2∗5+2√52−1−√5−1 | Simplify. |
| f(5)=√10+2√24−2 | Let's try to do the same process like above to simplify the radical: |
| √10+2√24 | |
| √10+2√24 | Rearrange the terms to put them in the form of a^2+2ab+b^2. Before we do that, however, we must simplify the square root of 24. |
| √10+4√6 | Now, do the same process as above |
| √10+4√6+√62−6 | Rearrange the terms and simplify. |
| √√62+4√6+4 | Yet again, this is a perfect square trinomial. Factor it. |
| √(√6+2)2 | Just like above, the square root and the square functions cancel. |
| √6+2 | Plug this in. |
| f(5)=√6+2−2 | |
| f(5)=√6 | |
Here's a third point (5,√6). Not every integer will be nice solutions. These, however, are examples of the nicest examples that you will get. After plotting these 3 points, connect them loosely like the shape of the parent function. I picked those points because I knew that they all would have a nice answer. Other points you plug in may not have as nice of a solution.
If you would like to reference a graph, I have one for you! Click here to view it:
+130466 Very crafty, X2 ....that one took a bit of work !!!!
I particularly like that trick of producing a perfect square from the sum of an integer and a radical multiplied by an integer !!!.....where did you learn that???
+2446 The trick does not always work, unfortunately. I just got very lucky that all of those examples actually simplify to something nice. Although I cannot locate where exactly I learned this awesome trick, I have located something that alters the method slightly, but it should give you an idea of what I am doing:
+2446 When I learned this trick, I was stunned, too. This cannot be done in every circumstance, however. I just got lucky that it happened to worked on every occasion. Although i cannot locate where I originally learned this method, I can provide you to a link where someone utilizes this method to simplify a complex radical expression. Here it is:
