# modulus questionWatch

#1
How do I do this?

Express

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

for the interval

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)?
0
7 years ago
#2
(Original post by Plato's Trousers)
How do I do this?

Express

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

for the interval

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

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.
1
#3
(Original post by dknt)
As you've correctly said the modulus will be positive in that given domain, so we can rewrite it as

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.

now, the bit in the brackets is only positive for part of the domain...
0
7 years ago
#4
(Original post by Plato's Trousers)
ok, thanks for that. I see now.

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
you do the same thing i.e.
0
7 years ago
#5
(Original post by Plato's Trousers)
ok, thanks for that. I see now.

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.
1
#6
(Original post by dknt)
So if they said find the linear function of x, with the domain
you do the same thing i.e.
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
0
7 years ago
#7
(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.
0
#8
thanks both (repped)
0
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