modulus question Watch

Plato's Trousers
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#1
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How do I do this?

Express y=x+|2x-1|

and a linear function of x in the form y=mx +c

for the interval x>\frac{1}{2}

I can see that over that interval, the mod part will be positive, but I can't see what they mean? Presumably the equation just becomes y=x+c (where c is positive)?
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dknt
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(Original post by Plato's Trousers)
How do I do this?

Express y=x+|2x-1|

and a linear function of x in the form y=mx +c

for the interval x>\frac{1}{2}

I can see that over that interval, the mod part will be positive, but I can't see what they mean? Presumably the equation just becomes y=x+c (where c is positive)?
As you've correctly said the modulus will be positive in that given domain, so we can rewrite it as y= x+[+(2x-1)]

Then it's just a simple case of removing brackets. If you imagine the graph, it's like a wonky "v" shape, so it's like it's made of 2 linear functions of x "glued" together, and it's asking you to find the equation for the positive arm.
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Plato's Trousers
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(Original post by dknt)
As you've correctly said the modulus will be positive in that given domain, so we can rewrite it as y= x+[+(2x-1)]

Then it's just a simple case of removing brackets. If you imagine the graph, it's like a wonky "v" shape, so it's like it's made of 2 linear functions of x "glued" together, and it's asking you to find the equation for the positive arm.
ok, thanks for that. I see now.

Just out of interest, what if they had asked the same question, but about
y=x+|2x-3|

now, the bit in the brackets is only positive for part of the domain...
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dknt
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(Original post by Plato's Trousers)
ok, thanks for that. I see now.

Just out of interest, what if they had asked the same question, but about
y=x+|2x-3|

now, the bit in the brackets is only positive for part of the domain...
So if they said find the linear function of x, with the domain x>\frac{3}{2}
you do the same thing i.e.  y= x+[+(2x-3)]
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Zuzuzu
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(Original post by Plato's Trousers)
ok, thanks for that. I see now.

Just out of interest, what if they had asked the same question, but about
y=x+|2x-3|

now, the bit in the brackets is only positive for part of the domain...
As far as I know, for those conditions it can't be expressed as y = mx + c.
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Plato's Trousers
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(Original post by dknt)
So if they said find the linear function of x, with the domain x>\frac{3}{2}
you do the same thing i.e.  y= x+[+(2x-3)]
no, i was meaning the question was the same apart from the function given. ie, the domain is still x>0.5

But I think zuzuzu is right, it can't be expressed as a linear function
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dknt
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(Original post by Plato's Trousers)
no, i was meaning the question was the same apart from the function given. ie, the domain is still x>0.5

But I think zuzuzu is right, it can't be expressed as a linear function
Ah right sorry, yeh, since below 3/2, the values in the modulus are negative, so you'll have part of the negative arm and the positive arm.
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Plato's Trousers
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thanks both (repped)
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