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modulus question

How do I do this?

Express y=x+2x1y=x+|2x-1|

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

for the interval x>12x>\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)?
Reply 1
Original post by Plato's Trousers
How do I do this?

Express y=x+2x1y=x+|2x-1|

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

for the interval x>12x>\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+[+(2x1)]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.
(edited 13 years ago)
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+[+(2x1)]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+2x3y=x+|2x-3|

now, the bit in the brackets is only positive for part of the domain...
Reply 3
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+2x3y=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>32x>\frac{3}{2}
you do the same thing i.e. y=x+[+(2x3)] y= x+[+(2x-3)]
Reply 4
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+2x3y=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.
Original post by dknt
So if they said find the linear function of x, with the domain x>32x>\frac{3}{2}
you do the same thing i.e. y=x+[+(2x3)] 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
Reply 6
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.
thanks both (repped)

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