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# graphs

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how to sketch the graph of y = sqrt(2x+2sqrt(x^2-1)) - sqrt(x-1) by hands

Aug 7, 2017

#1
+2340
+1

You are wondering how to graph the following equation of $$y=\sqrt{2x+2\sqrt{x^2-1}}-\sqrt{x-1}$$. First, we should determine the domain to see what the real inputs are:

 $$y=\sqrt{2x+2\sqrt{x^2-1}}-\sqrt{x-1}$$ Factor out a 2 from 2x+2 $$y=\sqrt{2(x+1)\sqrt{x^2-1}}-\sqrt{x-1}$$ Also, x^2-1 can be factored into (x+1)(x-1). $$y=\sqrt{2(x+1)\sqrt{(x+1)(x-1)}}-\sqrt{x-1}$$ Split the 2 out of the radical. $$y=\sqrt{2}\sqrt{x+\sqrt{(x+1)(x-1)}}-\sqrt{x-1}$$

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.
 $$x+1<0$$ $$x-1<0$$

Add or subtract 1 from both sides.
 $$x<-1$$ $$x<1$$

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:

$$\text{Domain:}\{\hspace{1mm}f(x)\in\mathbb{R}|x\geq1\}$$

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 $$f(x)=\sqrt{x}$$. 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:

 $$f(x)=\sqrt{2x+2\sqrt{x^2-1}}-\sqrt{x-1}$$ Replace every instance of x with one. $$f(1)=\sqrt{2*1+2\sqrt{1^2-1}}-\sqrt{1-1}$$ Simplify. $$f(1)=\sqrt{2+2\sqrt{0}}-\sqrt{0}$$ Of course, $$\sqrt{0}=0$$ $$f(1)=\sqrt{2}$$

Therefore, one point on the graph is $$(1,\sqrt{2})$$. Let's find another. Let's plug in 2:

 $$f(x)=\sqrt{2x+2\sqrt{x^2-1}}-\sqrt{x-1}$$ Replace every instance of x with two. $$f(2)=\sqrt{2*2+2\sqrt{2^2-1}}-\sqrt{2-1}$$ Simplify $$f(2)=\sqrt{4+2\sqrt{3}}-\sqrt{1}$$ $$\sqrt{1}=1$$, of course. Now, let's simplify this monstrosity$$\sqrt{4+2\sqrt{3}}$$ $$\sqrt{4+2\sqrt{3}+\sqrt{3}^2-3}$$ 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. $$\sqrt{\sqrt{3}^2+2\sqrt{3}+1}$$ It might be difficult to notice this, but the radicand is actually a perfect-square trinomial. $$\sqrt{(\sqrt{3}+1)^2}$$ The square root and the square function undo each other. $$\sqrt{3}+1$$ Now, let's replace $$\sqrt{4+2\sqrt{3}}$$ with its simplified value. $$f(2)=\sqrt{3}+1-1$$ $$f(2)=\sqrt{3}$$

Therefore, another point is $$(2,\sqrt{3})$$. Let's one more example. I know ahead of time that x=5 is a good point to plug in:

 $$f(x)=\sqrt{2x+2\sqrt{x^2-1}}-\sqrt{x-1}$$ Just like before, substitute x for 5. $$f(5)=\sqrt{2*5+2\sqrt{5^2-1}}-\sqrt{5-1}$$ Simplify. $$f(5)=\sqrt{10+2\sqrt{24}}-2$$ Let's try to do the same process like above to simplify the radical: $$\sqrt{10+2\sqrt{24}}$$ $$\sqrt{10+2\sqrt{24}}$$ 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. $$\sqrt{10+4\sqrt{6}}$$ Now, do the same process as above $$\sqrt{10+4\sqrt{6}+\sqrt{6}^2-6}$$ Rearrange the terms and simplify. $$\sqrt{\sqrt{6}^2+4\sqrt{6}+4}$$ Yet again, this is a perfect square trinomial. Factor it. $$\sqrt{(\sqrt{6+2})^2}$$ Just like above, the square root and the square functions cancel. $$\sqrt{6}+2$$ Plug this in. $$f(5)=\sqrt{6}+2-2$$ $$f(5)=\sqrt{6}$$

Here's a third point $$(5,\sqrt{6})$$. 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:

Aug 7, 2017
edited by TheXSquaredFactor  Aug 7, 2017

#1
+2340
+1

You are wondering how to graph the following equation of $$y=\sqrt{2x+2\sqrt{x^2-1}}-\sqrt{x-1}$$. First, we should determine the domain to see what the real inputs are:

 $$y=\sqrt{2x+2\sqrt{x^2-1}}-\sqrt{x-1}$$ Factor out a 2 from 2x+2 $$y=\sqrt{2(x+1)\sqrt{x^2-1}}-\sqrt{x-1}$$ Also, x^2-1 can be factored into (x+1)(x-1). $$y=\sqrt{2(x+1)\sqrt{(x+1)(x-1)}}-\sqrt{x-1}$$ Split the 2 out of the radical. $$y=\sqrt{2}\sqrt{x+\sqrt{(x+1)(x-1)}}-\sqrt{x-1}$$

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.
 $$x+1<0$$ $$x-1<0$$

Add or subtract 1 from both sides.
 $$x<-1$$ $$x<1$$

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:

$$\text{Domain:}\{\hspace{1mm}f(x)\in\mathbb{R}|x\geq1\}$$

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 $$f(x)=\sqrt{x}$$. 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:

 $$f(x)=\sqrt{2x+2\sqrt{x^2-1}}-\sqrt{x-1}$$ Replace every instance of x with one. $$f(1)=\sqrt{2*1+2\sqrt{1^2-1}}-\sqrt{1-1}$$ Simplify. $$f(1)=\sqrt{2+2\sqrt{0}}-\sqrt{0}$$ Of course, $$\sqrt{0}=0$$ $$f(1)=\sqrt{2}$$

Therefore, one point on the graph is $$(1,\sqrt{2})$$. Let's find another. Let's plug in 2:

 $$f(x)=\sqrt{2x+2\sqrt{x^2-1}}-\sqrt{x-1}$$ Replace every instance of x with two. $$f(2)=\sqrt{2*2+2\sqrt{2^2-1}}-\sqrt{2-1}$$ Simplify $$f(2)=\sqrt{4+2\sqrt{3}}-\sqrt{1}$$ $$\sqrt{1}=1$$, of course. Now, let's simplify this monstrosity$$\sqrt{4+2\sqrt{3}}$$ $$\sqrt{4+2\sqrt{3}+\sqrt{3}^2-3}$$ 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. $$\sqrt{\sqrt{3}^2+2\sqrt{3}+1}$$ It might be difficult to notice this, but the radicand is actually a perfect-square trinomial. $$\sqrt{(\sqrt{3}+1)^2}$$ The square root and the square function undo each other. $$\sqrt{3}+1$$ Now, let's replace $$\sqrt{4+2\sqrt{3}}$$ with its simplified value. $$f(2)=\sqrt{3}+1-1$$ $$f(2)=\sqrt{3}$$

Therefore, another point is $$(2,\sqrt{3})$$. Let's one more example. I know ahead of time that x=5 is a good point to plug in:

 $$f(x)=\sqrt{2x+2\sqrt{x^2-1}}-\sqrt{x-1}$$ Just like before, substitute x for 5. $$f(5)=\sqrt{2*5+2\sqrt{5^2-1}}-\sqrt{5-1}$$ Simplify. $$f(5)=\sqrt{10+2\sqrt{24}}-2$$ Let's try to do the same process like above to simplify the radical: $$\sqrt{10+2\sqrt{24}}$$ $$\sqrt{10+2\sqrt{24}}$$ 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. $$\sqrt{10+4\sqrt{6}}$$ Now, do the same process as above $$\sqrt{10+4\sqrt{6}+\sqrt{6}^2-6}$$ Rearrange the terms and simplify. $$\sqrt{\sqrt{6}^2+4\sqrt{6}+4}$$ Yet again, this is a perfect square trinomial. Factor it. $$\sqrt{(\sqrt{6+2})^2}$$ Just like above, the square root and the square functions cancel. $$\sqrt{6}+2$$ Plug this in. $$f(5)=\sqrt{6}+2-2$$ $$f(5)=\sqrt{6}$$

Here's a third point $$(5,\sqrt{6})$$. 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:

TheXSquaredFactor Aug 7, 2017
edited by TheXSquaredFactor  Aug 7, 2017
#2
+98129
0

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???

Aug 7, 2017
edited by CPhill  Aug 7, 2017
#3
+2340
0

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:

TheXSquaredFactor  Aug 7, 2017
#4
+2340
0

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:

TheXSquaredFactor  Aug 8, 2017