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    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
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    I'm sure people would be willing to help you, without the clickbait...
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    (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"?
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    (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?
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    (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
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    (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
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    (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.
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    (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
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    (Original post by Katiee224)
    aha don't worry I'll figure it out

    thanks for the help
    No worries.
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    Goddammit I came here to see how :cry2:
 
 
 
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