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    Hey, I was just confused on how to actually do these two questions.
    I did number 7a (I got 3x^2 + 12x +12) but not the other three. I would appreciate some help. Thanks! 😀

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    (Original post by homeland.lsw)
    Hey, I was just confused on how to actually do these two questions.
    I did number 7a (I got 3x^2 + 12x +12) but not the other three. I would appreciate some help. Thanks! 😀

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    Lovely work on part(a), it's all correct!

    For part(b) you know that stationary points are solutions to the equation \frac{\mathrm{d}y}{\mathrm{d}x} = 0, can you solve this equation and find the x-coordinates of the stationary points?
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    (Original post by Zacken)
    Lovely work on part(a), it's all correct!

    For part(b) you know that stationary points are solutions to the equation \frac{\mathrm{d}y}{\mathrm{d}x} = 0, can you solve this equation and find the x-coordinates of the stationary points?
    So would I factorise the answer to part a to get 3(x+2)(x+2) and then solve for x?
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    (Original post by homeland.lsw)
    Hey, I was just confused on how to actually do these two questions.
    I did number 7a (I got 3x^2 + 12x +12) but not the other three. I would appreciate some help. Thanks! 😀

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    For 7.b., you need to remember the important fact that \frac{dy}{dx}=0 at stationary points and that \frac{d^2y}{dx^2}\leq 0 at maxima (i.e. there is downwards curvature, the gradient is decreasing) and \frac{d^2y}{dx^2}\geq 0 at minima (i.e. there is upwards curvature, the gradient is increasing).

    This should help you answer question 8.b. too. For 8.c, think about the shape of inflection points - how might you show it's an inflection point? Hint: you may want to use \frac{dy}{dx}.
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    (Original post by homeland.lsw)
    Hey, I was just confused on how to actually do these two questions.
    I did number 7a (I got 3x^2 + 12x +12) but not the other three. I would appreciate some help. Thanks! 😀

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    For 7b) Derive the equation again, as it is a stationary point d2y/dx2 must =0, rearrange for x and sub it back into the orignal equation to get they y coordinate
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    (Original post by homeland.lsw)
    So would I factorise the answer to part a to get 3(x+2)(x+2) and then solve for x?
    Yes, that's correct. Once you've found the requisite values of x you then need to check whether \frac{d^2y}{dx^2} is positive, negative or zero for each of the x's that you've found to categorise it as a minimum, maximum or inflection.
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    (Original post by K-Fox)
    For 7b) Derive the equation again,
    Do you mean differentiate?
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    (Original post by Zacken)
    Yes, that's correct. Once you've found the requisite values of x you then need to check whether \frac{d^2y}{dx^2} is positive, negative or zero for each of the x's that you've found to categorise it as a minimum, maximum or inflection.
    Ok great, so my answer is this...
    Attachment 511841511843511900

    Also what's the story with that \frac{d^2y}{dx^2}?It's a bit confusing...

    Edit I don't know why there are three attachments
    Attached Images
       
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    (Original post by homeland.lsw)
    Ok great, so my answer is this...
    Attachment 511841511843511900

    Also what's the story with that \frac{d^2y}{dx^2}?It's a bit confusing...

    Edit I don't know why there are three attachments
    \frac{d^2y}{dx^2}= \frac{d}{dx} \frac{dy}{dx} = \frac{d}{dx} (3x^2 + 12x +12) have you seen this before? It's the derivative of the derivative.
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    (Original post by Zacken)
    \frac{d^2y}{dx^2}= \frac{d}{dx} \frac{dy}{dx} = \frac{d}{dx} (3x^2 + 12x +12) have you seen this before? It's the derivative of the derivative.
    I haven't no...we covered maxima and minima using the method from question 7 and a table...

    Anyway my question 7b is correct I presume? And I'll get on with question 8.
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    (Original post by homeland.lsw)
    Ok great, so my answer is this...
    Oh, and your stationary points is perfect. Good work! You just need to determine whether it's a minimum, maximum or inflection point now.

    Once you find your second derivative (by differentiating your first derivative) you should get an equation that when you plug x=-2 in, it'll either give you a positive value (x=-2 is a minimum), a negative value (x=-2 is a maximum) or a zero (x=-2 is an inflection point).
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    (Original post by Zacken)
    Oh, and your stationary points is perfect. Good work! You just need to determine whether it's a minimum, maximum or inflection point now.

    Once you find your second derivative (by differentiating your first derivative) you should get an equation that when you plug x=-2 in, it'll either give you a positive value (x=-2 is a minimum), a negative value (x=-2 is a maximum) or a zero (x=-2 is an inflection point).
    And if by magic it gave me zero!!!

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    Also you remind me of my maths teacher by saying "plug in" :lol:
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    (Original post by homeland.lsw)
    And if by magic it gave me zero!!!

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    Also you remind me of my maths teacher by saying "plug in" :lol:
    Yeees! Inflection it is. I get the same thing.

    I really should be saying "substitute" instead of "plug in". :lol:
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    (Original post by Zacken)
    Yeees! Inflection it is. I get the same thing.

    I really should be saying "substitute" instead of "plug in". :lol:
    Ok so this is my number 8...
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    Attachment 511857511859
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    (Original post by homeland.lsw)
    Ok so this is my number 8...
    Attachment 511857511859
    All correct. :yep:
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    (Original post by Zacken)
    All correct. :yep:
    I'm practically a mathematician now!! :lol:
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    (Original post by homeland.lsw)
    I'm practically a mathematician now!! :lol:
    I concur.
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    (Original post by longshot100)
    ...
    He's gotten his answer already.
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    (Original post by Zacken)
    He's gotten his answer already.
    Oh lol, sorry
 
 
 
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