+0  
 
0
416
4
avatar

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

 Aug 7, 2017

Best Answer 

 #1
avatar+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:

Source: https://dadkins.wikispaces.com/file/view/f%28x%29%3Dsquare_root_of_x.png/268949096/383x298/f%28x%29%3Dsquare_root_of_x.png

 

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
avatar+2340 
+1
Best Answer

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:

Source: https://dadkins.wikispaces.com/file/view/f%28x%29%3Dsquare_root_of_x.png/268949096/383x298/f%28x%29%3Dsquare_root_of_x.png

 

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
avatar+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???

 

 

cool cool cool

 Aug 7, 2017
edited by CPhill  Aug 7, 2017
 #3
avatar+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:

 

https://www.youtube.com/watch?v=naGkELTaSuQ&t=39s

TheXSquaredFactor  Aug 7, 2017
 #4
avatar+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:

 

https://www.youtube.com/watch?v=naGkELTaSuQ&t=77s

TheXSquaredFactor  Aug 8, 2017

12 Online Users

avatar
avatar
avatar