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    (Original post by jacksonmeg)
    dont you put in the x values and the positive one is minimum, negative is max?
    Essentially yes.

    The description of the test
    When the value associated with your point is substituted into the second derivative, the value yielded will tell us if the point is a maximum, minimum, or the test is inconclusive which will require the evaluation of the third derivative.

    If \ \dfrac{d^{2}y}{dx^{2}} < 0 \ the \ point \ is \ a \ local \ maximum

    If \ \dfrac{d^{2}y}{dx^{2}} > 0 \ the \ point \ is \ a \ local \ minimum

    If \ \dfrac{d^{2}y}{dx^{2}} = 0 \ the \ test \ is \ inconclusive

    We now know that:

    \dfrac{dy}{dx} \Big|_{x=0} = 250

    \dfrac{dy}{dx} \Big|_{x=2} = -90

    \dfrac{dy}{dx} \Big|_{x=5} = 0

    So you can now deduce which is the maximum and which is the minimum.

    (Original post by jacksonmeg)
    I figured it didnt matter about simplifying since you just need to put in x values, and i usually go wrong anyway. lol
    I'm the exact same. That's why I got into the habit of simplifying everything so I wouldn't make any silly errors!
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    (Original post by Khallil)
    Essentially yes.

    The description of the test
    When the value associated with your point is substituted into the second derivative, the value yielded will tell us if the point is a maximum, minimum, or the test is inconclusive which will require the evaluation of the third derivative.

    If \ \dfrac{d^{2}y}{dx^{2}} < 0 \ the \ point \ is \ a \ local \ maximum

    If \ \dfrac{d^{2}y}{dx^{2}} > 0 \ the \ point \ is \ a \ local \ minimum

    If \ \dfrac{d^{2}y}{dx^{2}} = 0 \ the \ test \ is \ inconclusive

    We now know that:

    \dfrac{dy}{dx} \Big|_{x=0} = 250

    \dfrac{dy}{dx} \Big|_{x=2} = -90

    \dfrac{dy}{dx} \Big|_{x=5} = 0

    So you can now deduce which is the maximum and which is the minimum.



    I'm the exact same. That's why I got into the habit of simplifying everything so I wouldn't make any silly errors!
    I was told the numbers had to be the same, e.g. +9 and -9, is this wrong?
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    (Original post by jacksonmeg)
    I was told the numbers had to be the same, e.g. +9 and -9, is this wrong?
    The values for the second derivative?
    If that's the case, then no. They don't have to be the same at all.

    Whoever told you this must have been having a bad day.
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    (Original post by BankOfPigs)
    Think it's a bit easier to do u= (5-x)^2 v = -5x^2+10x. Then you don't end up using the product rule twice.
    We know that dy/dx = x(5-x)^2 [10 -5x]
    Change this to: (5-x)^2 [10x-5x^2] (just multiplying in the x)
    Now make u= (5-x)^2 and v= 10x-5x^2
    du/dx = -2(5-x) and dv/dx = 10 -10x
    Using the product rule we get
    d^2 y / dx^2 = (10x-5x^2)(-10+2x) + (5-x)^2(10-10x)
    Substitute in your values that you found for the turning points (x=0,5,2) into the above equation.

    Can you go from here?
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    (Original post by Khallil)
    x
    Is my answer correct above? I ask as your answer seems very different to my answer, (I help people on TSR as a form of revision for myself).
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    (Original post by Alex-Torres)
    Is my answer correct above? I ask as your answer seems very different to my answer, (I help people on TSR as a form of revision for myself).
    They're both the same. I just factored out 10 and (5-x).

    \begin{aligned} \dfrac{d^2y}{dx^2} & = \text{Your answer} \\ & = (10x-5x^2)(-10+2x) + (5-x)^2(10-10x) \\ & = 5(2x-x^2)(-10+2x) + 10(5-x)^2(1-x) \\ & = -10(2x-x^2)(5-x) + 10(5-x)^2(1-x) \\ & = 10(5-x) \left( -(2x-x^2) + (5-x)(1-x) \right) \\ & = 10(5-x) \left( x^2 - 2x + 5 - 6x + x^2 \right) \\ & = 10(5-x) \left( 2x^2 - 8x + 5 \right) \\ & = \text{My answer} \end{aligned}

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    (Original post by Khallil)
    They're both the same. I just factored out 10 and (5-x).

    \begin{aligned} \dfrac{d^2y}{dx^2} & = \text{Your answer} \\ & = (10x-5x^2)(-10+2x) + (5-x)^2(10-10x) \\ & = 5(2x-x^2)(-10+2x) + 10(5-x)^2(1-x) \\ & = -10(2x-x^2)(5-x) + 10(5-x)^2(1-x) \\ & = 10(5-x) \left( -(2x-x^2) + (5-x)(1-x) \right) \\ & = 10(5-x) \left( x^2 - 2x + 5 - 6x + x^2 \right) \\ & = 10(5-x) \left( 2x^2 - 8x + 5 \right) \\ & = \text{My answer} \end{aligned}

    right, so 250 is the min, -90 the max?
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    (Original post by Alex-Torres)
    Is my answer correct above? I ask as your answer seems very different to my answer, (I help people on TSR as a form of revision for myself).
    Could you check this for me please?

    Solve |x-2| < 2|x-1|
    square both sides
    (x-2)(x-2) < 2(x-1)(x-1)
    x^2 -4x +4 < 2(x^2 -2x +1)
    x^2 -4x +4 < 2x^2 -4x +2
    0 < x^2 -2
    2 < x^2
    -sqrt(2) < x < sqrt(2)
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    (Original post by jacksonmeg)
    Could you check this for me please?

    Solve |x-2| < 2|x-1|
    square both sides
    (x-2)(x-2) < 2(x-1)(x-1)
    x^2 -4x +4 < 2(x^2 -2x +1)
    x^2 -4x +4 < 2x^2 -4x +2
    0 < x^2 -2
    2 < x^2
    -sqrt(2) < x < sqrt(2)
    Firstly, do you know what the answer is?
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    (Original post by Alex-Torres)
    Firstly, do you know what the answer is?
    no
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    (Original post by jacksonmeg)
    no
    Try sketching both mod graphs and looking at the intervals in which one is greater than the other if you're struggling with the squaring (IMO also harder) method.
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    (Original post by jacksonmeg)
    right, so 250 is the min, -90 the max?
    The point at which the second derivative is equal to 250 is the local minimum.

    The point associated with a second derivative of -90 is a local maximum.

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    (Original post by jacksonmeg)
    Could you check this for me please?

    Solve |x-2| < 2|x-1|
    square both sides
    (x-2)(x-2) < 2(x-1)(x-1)
    x^2 -4x +4 < 2(x^2 -2x +1)
    x^2 -4x +4 < 2x^2 -4x +2
    0 < x^2 -2
    2 < x^2
    -sqrt(2) < x < sqrt(2)
    (Original post by Liamnut)
    Try sketching both mod graphs and looking at the intervals in which one is greater than the other if you're struggling with the squaring (IMO also harder) method.
    This is the best method for questions like these!
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    (Original post by Khallil)
    This is the best method for questions like these!
    im not sure what 2|x-1| looks like
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    (Original post by jacksonmeg)
    im not sure what 2|x-1| looks like
    This is just a basic transformation of the absolute value function.

    Hint
    If I told you that f(x) = |x|, f(x+a) is a horizontal translation to the left by a units and af(x) is a vertical stretch of scale factor a. What would the graph of 2|x-1| look like?
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    (Original post by Khallil)
    If I told you that f(x) = |x|, f(x+a) is a horizontal translation to the left by a units and af(x) is a vertical stretch of scale factor a. What would the graph of 2|x-1| look like?
    i mean, do you stretch x-1 by factor of 2 then take the modulus, or do the modulus of x-1 then stretch it? does it make a difference?
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    (Original post by jacksonmeg)
    i mean, do you stretch x-1 by factor of 2 then take the modulus, or do the modulus of x-1 then stretch it? does it make a difference?
    It doesn't make a difference in this case.
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    (Original post by jacksonmeg)
    i mean, do you stretch x-1 by factor of 2 then take the modulus, or do the modulus of x-1 then stretch it? does it make a difference?
    Note that 2|x-1| = |2x - 2| which may be an easier way to visualize it
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    (Original post by Khallil)
    It doesn't make a difference in this case.
    is one of them x<0
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    (Original post by jacksonmeg)
    is one of them x<0
    :yep:

    Now find the other one!
 
 
 

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