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    (Original post by Zacken)
    wot?

    for a stationary point to be an inflection point, you need f'(x) = 0 and f''(x) = 0.

    [and the obvious caveat for anybody who cares]
    This is what I dont get

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    (Original post by RDKGames)
    I'll have some of whatever drugs you're having, mate.

    Stationary points where f'(x)=0
    Points of inflection where f''(x)=0

    Combine the two for both to be true.

    "dy/dx must be positive/negative and NOT change its sign (-/+)" ...huh?
    "so what i do is the 'gradient table'..." ...why? Just solve for values of x then solve for values of x on f''(x)=0. Compare the two for the same x's and you've found the x coordinates of the points which are stationary AND points of inflection at the same time.
    I dont know tbh what is the example in my book on about

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    (Original post by ihatePE)
    Hello, i just want a confirmation relating to ''determining whether a stationary point is a point of inflection''

    when f''(x) = 0 you need to investigate further right?
    so what i do is the 'gradient table' where i find out what dy/dx is when x=-1 x=0 x=1 ect

    my point is, for it to be a point of inflection, dy/dx must be positive/negative and NOT change its sign (-/+) when x=-1 and x=1... am i right?

    i've read the notes in the book for ages and its worded so briefly that im not so sure if that's the rules.
    For a point to be an inflection point it is necessary that at that point  d^2y/dx^2=0 but all because  d^2y/dx^2=0 does not mean it is an inflection point necessarily.
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    (Original post by ihatePE)
    I dont know tbh what is the example in my book on about

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    Do you know what an inflection point actually is?
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    In particular, to back up B_9710's comment, you'll see that you need the third derivative to be non-zero. if you're not convinced, look at x^4.
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    (Original post by B_9710)
    Do you know what an inflection point actually is?
    it's when the gradient on either side of the stationary point is 0?
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    (Original post by ihatePE)
    it's when the gradient on either side of the stationary point is 0?
    read that back to yourself, what does it mean for the gradient to be 0?
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    (Original post by ihatePE)
    it's when the gradient on either side of the stationary point is 0?
    https://en.wikipedia.org/wiki/Inflection_point
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    (Original post by Zacken)
    read that back to yourself, what does it mean for the gradient to be 0?
    oh stationary... inflection is when it doesnt change on either side of the stationary point so like positive on left and positive on right/ neg on left and right? something like that ??
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    complete the square question

    a)express x^2 + 6x - 4 in the form (x+a)^2 + b
    b) use your results to part a) to find the least value of 2x^2 + 12x - 8 and the corresponding value of x


    i know how to do part a), but im not sure what part b) is even asking. i talked about this with my teacher, she said take the 2 out as the correspondant
    2 [ x^2 + 6x - 4 ]
    but i still dont get what she want me to do afterwards
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    (Original post by ihatePE)
    complete the square question

    a)express x^2 + 6x - 4 in the form (x+a)^2 + b
    b) use your results to part a) to find the least value of 2x^2 + 12x - 8 and the corresponding value of x


    i know how to do part a), but im not sure what part b) is even asking. i talked about this with my teacher, she said take the 2 out as the correspondant
    2 [ x^2 + 6x - 4 ]
    but i still dont get what she want me to do afterwards
    Okay, so you factored out the 2. Now put the completed square form inside the bracket. Expand (the outer bracket) and find when the value is a minimum. That will be when 2(x-3)^2=0 of course as the expression is always positive and anything other than 0 would put you above your minimum value.
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    (Original post by RDKGames)
    Okay, so you factored out the 2. Now put the completed square form inside the bracket. Expand (the outer bracket) and find when the value is a minimum. That will be when 2(x-3)^2=0 of course as the expression is always positive and anything other than 0 would put you above your minimum value.
    thanks
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    show that
    x^2 + (2k - 1)x + (k^2 - k + 2) = 0
    has no real roots, whatever the value of the constant k (4 marks)

    and find the range of values of x satisfying the inequality
    3x^2 + 16x - 12 > 0



    for the first part i got to

    ''for no real roots b^2 - 4ac < 0
    4k^2 + 1 - 4(k^2-k+2)
    -4k+9=0
    k=-9/-4
    hence k<0


    is that it? im confused at the wording of the question
    and the second quetion is part b) and usually when this happen, it links with the first part of the question, i just want to know if there's a link? if not then i dont need help with the second question, i can do that
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    (Original post by ihatePE)
    show that
    x^2 + (2k - 1)x + (k^2 - k + 2) = 0
    has no real roots, whatever the value of the constant k (4 marks)

    and find the range of values of x satisfying the inequality
    3x^2 + 16x - 12 > 0



    for the first part i got to

    ''for no real roots b^2 - 4ac < 0
    4k^2 + 1 - 4(k^2-k+2)
    -4k+9=0
    k=-9/-4
    hence k<0


    is that it? im confused at the wording of the question
    and the second quetion is part b) and usually when this happen, it links with the first part of the question, i just want to know if there's a link? if not then i dont need help with the second question, i can do that
    You squared the bracket wrong and be careful with the signs, for there to be no real roots the discriminant has to be negative; there's no k terms in the discriminant.

    For the second part, factorise and find the critical values.
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    BINOMIAL EXPANSION help!
    my text book says to use the formula


    but my teacher taught me the pascal triangle which seems like less work. im just wondering if they both do the same thing, no more no less? i prefer the triangle by the looks and i really dont want to have to work with this awful long winded formula
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    (Original post by ihatePE)
    BINOMIAL EXPANSION help!
    my text book says to use the formula


    but my teacher taught me the pascal triangle which seems like less work. im just wondering if they both do the same thing, no more no less? i prefer the triangle by the looks and i really dont want to have to work with this awful long winded formula
    You can't use the Triangle for everything, What if it is to the power of 8 or something like that will you draw out then. You don't have to memorise the formula learn the method and practice it often, it will become easier then.
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    (Original post by ihatePE)
    BINOMIAL EXPANSION help!
    my text book says to use the formula


    but my teacher taught me the pascal triangle which seems like less work. im just wondering if they both do the same thing, no more no less? i prefer the triangle by the looks and i really dont want to have to work with this awful long winded formula
    Sure you can use the triangle but the triangle doesnt make it very clear when dealing with something like (3+2x)^4


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    (Original post by TheAlphaParticle)
    You can't use the Triangle for everything, What if it is to the power of 8 or something like that will you draw out then. You don't have to memorise the formula learn the method and practice it often, it will become easier then.
    my teacher said my exam board ''usually' like to stay in the range of powers of 5 :hoppy: but i'll use the formula anyways just to get the hang of using both. i'll ask my teacher for her opinion tomorrow as well.


    (Original post by RDKGames)
    Sure you can use the triangle but the triangle doesnt make it very clear when dealing with something like (3+2x)^4


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    mmmm is that a challenge? :sexface:
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    (Original post by ihatePE)
    my teacher said my exam board ''usually' like to stay in the range of powers of 5 :hoppy: but i'll use the formula anyways just to get the hang of using both. i'll ask my teacher for her opinion tomorrow as well.



    mmmm is that a challenge? :sexface:
    Lol go for it - expand it using the triangle then


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    (Original post by RDKGames)
    Lol go for it - expand it using the triangle then


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    im quite tired so i did most of the calc in my head but this is what i have

    81 + 176x + 216x^2 + 106^3 + 16x^4
 
 
 
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