Hi guys,
I'm really struggling to understand how to sketch the modulus graphs and also use that graph to solve the equations . in the example blow part a, should i just sketch the graph of 2x+1 first and then reflect the negative yvalues in xaxis?

Alen.m
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 12062016 09:45

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 12062016 09:54
(Original post by Alen.m)
Hi guys,
I'm really struggling to understand how to sketch the modulus graphs and also use that graph to solve the equations . in the example blow part a, should i just sketch the graph of 2x+1 first and then reflect the negative yvalues in xaxis? 
Alen.m
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 12062016 10:20
(Original post by 13 1 20 8 42)
You can ignore the 4 and the minus first, and just think about the graph 2x + 1. This graph is quite simple: for all x < 1/2 you will have that 2x + 1 is negative, right? And for all x > 1/2, 2x + 1 is nonnegative. Hence their will be a "cusp" where x = 1/2, i.e. you reflect in the line x = 1/2. Then to obtain the new graph all you have to do is reflect that in the y axis, since its minus, and translate it, to account for the +4. 
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 12062016 10:22
(Original post by Alen.m)
do you think my way of dealing with this is also correct as well?first sketch the graph of 2x+1 and then reflect the negative yvalus in the xaxis to get 2x+1, the multiply ycoordinates of any turning point of the graph by 1 to get 2x+1 and at the end just translate by 4 vertically
Well I mean you multiply everything by 1 I'm not sure what you mean by multiplying the coordinates of the turning point 
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 12062016 10:30
(Original post by 13 1 20 8 42)
oops I meant reflect in the xaxis in my previous post by the way. Yes, that is right I think
Well I mean you multiply everything by 1 I'm not sure what you mean by multiplying the coordinates of the turning point 
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 12062016 10:33
(Original post by Alen.m)
Like every single point on the graph should be multiplied by 1? 
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 12062016 10:59
(Original post by 13 1 20 8 42)
Well yeah once you've got 2x + 1, to get 2x +1, you flip everything in the xaxis. And then add the 4 to get the final graph 
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 12062016 11:03
(Original post by Alen.m)
how about solving the equation in part b?im not sure what method would be the best to use as i keep getting wrong answers for part b
for x >/= 0, x = x
for x < 0, x = x
And you can do the same thing for 2x + 1, where the critical point is x = 1/2 rather than x = 0. Then, depending on the value of x, you have an explicit, direct equation to solve, without moduli in the way. You just need to check that whatever answer you get is actually in the range you're considering.
So for an example, if you had 3  x = 2, you'd say first look at x >/= 0, this gives the equation 3  x = 2 giving the solution x = 1, which is in the range you're considering, so is a solution. Then you'd look at x < 0, giving the equation 3 + x = 2, yielding x = 1, which again is in the range, so is a solution. This problem is no different, just the expression is a tiny bit more complicated.
(by the way to clarify I know that is a dumb example because you can rearrange, but I use it to illustrate a simple case..)Last edited by math42; 12062016 at 11:06. 
Alen.m
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 12062016 11:11
(Original post by 13 1 20 8 42)
Do it by case analysis. Always think of x as a function made up of pieces
for x >/= 0, x = x
for x < 0, x = x
And you can do the same thing for 2x + 1, where the critical point is x = 1/2 rather than x = 0. Then, depending on the value of x, you have an explicit, direct equation to solve, without moduli in the way. You just need to check that whatever answer you get is actually in the range you're considering.
So for an example, if you had 3  x = 2, you'd say first look at x >/= 0, this gives the equation 3  x = 2 giving the solution x = 1, which is in the range you're considering, so is a solution. Then you'd look at x < 0, giving the equation 3 + x = 2, yielding x = 1, which again is in the range, so is a solution. This problem is no different, just the expression is a tiny bit more complicated.
(by the way to clarify I know that is a dumb example because you can rearrange, but I use it to illustrate a simple case..) 
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 12062016 11:13
(Original post by Alen.m)
so should i start by considering x>1.5 and x<1.5? 
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 12062016 11:16
(Original post by 13 1 20 8 42)
well the important value here is x = 1/2 as this is where 2x + 1 = 0. 
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 12062016 11:21
(Original post by Alen.m)
so for x>1.5 we have 2x+1=4 to solve and for x<1.5 we have 2x1=4 to solve ?
x = 4  (2x + 1)
or
x = 4 + (2x + 1)
depending on which side of 1/2 you are on 
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 12062016 11:37
(Original post by 13 1 20 8 42)
well it looks like you're trying to solve for RHS = 0 there (also as I said we should be looking at 1/2 not 1.5). You have either
x = 4  (2x + 1)
or
x = 4 + (2x + 1)
depending on which side of 1/2 you are on 
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 12062016 11:40
(Original post by Alen.m)
Oh sorry i meant to say 0.5 . so we've got x=1.5 and x=1.5 as our two solutions? 
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 12062016 11:49
(Original post by 13 1 20 8 42)
solutions I get are 1 and 5 
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 12062016 11:51
(Original post by Alen.m)
Dont you have to rearrange the two above equations to find the value of x? 
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 12062016 13:44
(Original post by 13 1 20 8 42)
yes, I don't see where you get x = +/1.5 from though 
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 12062016 13:49
(Original post by Alen.m)
Move 4 and 1 to the other side and then devide by 2 to make x the subject?
x = 4 + (2x + 1) gives x = 5 + 2x so rearranging x = 5 
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 12062016 15:11
(Original post by 13 1 20 8 42)
x = 4  (2x + 1) gives x = 4  2x  1 so rearranging 3x = 3 hence x = 1
x = 4 + (2x + 1) gives x = 5 + 2x so rearranging x = 5 
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 12062016 15:14
(Original post by Alen.m)
My mistake thanks for you time
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