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# Fermats Last Theorem is false!

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1. Now that I've got your attention, can someone please tell me what's going in in part d of q4?

I know it's not complicated but I don't understand what they want to see

thanks

QP: https://a086a5a2f39bda93734c56a63fab...%20Edexcel.pdf

Zacken

2. I'm sure people would be willing to help you, without the clickbait...
3. (Original post by Katiee224)
Now that I've got your attention, can someone please tell me what's going in in part d of q4?

I know it's not complicated but I don't understand what they want to see

thanks

QP: https://a086a5a2f39bda93734c56a63fab...%20Edexcel.pdf

Zacken
f(x) = k means that the line y = k has to intersect the graph of f(x), for there to be two roots, it has to intersect twice. Look at your graph, for which values of k will the line y=k cut the graph of f(x) twice? There'll be no roots beyond that minimum corner of f(x), and only 1 above a certain value of k, but between those two values, there'll be two roots. What are those "between two values"?
4. (Original post by Zacken)
f(x) = k means that the line y = k has to intersect the graph of f(x), for there to be two roots, it has to intersect twice. Look at your graph, for which values of k will the line y=k cut the graph of f(x) twice? There'll be no roots beyond that minimum corner of f(x), and only 1 above a certain value of k, but between those two values, there'll be two roots. What are those "between two values"?
okay so f(0) = 11 so that's the upper limit for k

do I do f'(x) = 0 for the min point and that's the lower limit for k?

Also do I treat the modulus brackets as just normal brackets to do this?
5. (Original post by Katiee224)
okay so f(0) = 11 so that's the upper limit for k

do I do f'(x) = 0 for the min point and that's the lower limit for k?

Also do I treat the modulus brackets as just normal brackets to do this?
First bit is right. Second bit isn't. You're over complicating it.

f(x) is at an absolute minimum when x = 3 (you should be able to spot this, since you want to make the mod as small as possible to make f(x) as small as possible) so f(3) = 5
6. (Original post by Zacken)
First bit is right. Second bit isn't. You're over complicating it.

f(x) is at an absolute minimum when x = 3 (you should be able to spot this, since you want to make the mod as small as possible to make f(x) as small as possible) so f(3) = 5
okay I get it now thank you

another thing I'm not sure on, If i subbed in an x-value into the non-modulus part of the graph do I leave the brackets as (3-x)?

And If I am subbing an x-value which is in the modulus part of the graph, use -(3-x)? (like in fp2)

if that doesn't make sense don't worry hehe
7. (Original post by Katiee224)
okay I get it now thank you

another thing I'm not sure on, If i subbed in an x-value into the non-modulus part of the graph do I leave the brackets as (3-x)?

And If I am subbing an x-value which is in the modulus part of the graph, use -(3-x)? (like in fp2)

if that doesn't make sense don't worry hehe
I'm not sure I understand that, sorry.
8. (Original post by Zacken)
I'm not sure I understand that, sorry.
aha don't worry I'll figure it out

thanks for the help
9. (Original post by Katiee224)
aha don't worry I'll figure it out

thanks for the help
No worries.
10. Goddammit I came here to see how

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