Hollymae764
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Can someone explain to me how you find the cusp or peak of the modulus graph when you have numbers in front and behind the modulus.

Also could you help me with part b as I do not know how to inverse a modulus.
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RDKGames
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(Original post by Hollymae764)
Can someone explain to me how you find the cusp or peak of the modulus graph when you have numbers in front and behind the modulus.

Also could you help me with part b as I do not know how to inverse a modulus.
The cusp occurs when the stuff inside the modulus part is zero. In this question, this is when x+4 = 0.

Part b doesnt require you to find the inverse. It asks you to explain why it doesn't exist... Think about what kind of mapping a function needs to be in order for it to have an inverse.
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Hollymae764
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(Original post by RDKGames)
The cusp occurs when the stuff inside the modulus part is zero. In this question, this is when x+4 = 0.

Part b doesnt require you to find the inverse. It asks you to explain why it doesn't exist... Think about what kind of mapping a function needs to be in order for it to have an inverse.
so the cusp is still at x = -4
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Hollymae764
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so the range would be f(x) < 8

(Original post by Hollymae764)
so the cusp is still at x = -4
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Hollymae764
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(Original post by RDKGames)
The cusp occurs when the stuff inside the modulus part is zero. In this question, this is when x+4 = 0.

Part b doesnt require you to find the inverse. It asks you to explain why it doesn't exist... Think about what kind of mapping a function needs to be in order for it to have an inverse.
It would have to be reflected in y = x and since the function is a modulus this cannot happen. Is that correct?
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RDKGames
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(Original post by Hollymae764)
so the range would be f(x) < 8
Yes with \leq and not &lt;.

You do not need to use x=-4 here at all. Since the modulus is always non-negative, it means that

|x-4| \geq 0

so

-\dfrac{5}{3}|x-4| \leq 0

so

-\dfrac{5}{3}|x-4| + 8 \leq 8
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RDKGames
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(Original post by Hollymae764)
It would have to be reflected in y = x and since the function is a modulus this cannot happen. Is that correct?
That's not it. You have four mappings;

One-to-one
One-to-many
Many-to-one
Many-to-many

Which one can have an inverse? Which one does the modulus function correspond to?
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Hollymae764
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(Original post by RDKGames)
That's not it. You have four mappings;

One-to-one
One-to-many
Many-to-one
Many-to-many

Which one can have an inverse? Which one does the modulus function correspond to?
ohhh this is a many to one function and the inverse of it would be one to many which means it cant be a function
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RDKGames
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(Original post by Hollymae764)
ohhh this is a many to one function and the inverse of it would be one to many which means it cant be a function
Er.. close, one to many mappings are not functions to begin with, so no point trying to talk about inverse of this type of mapping.
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Hollymae764
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(Original post by RDKGames)
Er.. close, one to many mappings are not functions to begin with, so no point trying to talk about inverse of this type of mapping.
Im confused how is the modulus a one to many graph? I thought it was many to one?
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RDKGames
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(Original post by Hollymae764)
Im confused how is the modulus a one to many graph? I thought it was many to one?
Sorry I misread what you said. You're right in your reasoning.

More to the point, inverses only exist for one-to-one mappings but the modulus function is many-to-one.
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