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    Heya, could somebody check this question
    Find the coordinates of the maximum and minimum point on the curve y=x^2(5-x)^3, I got (5,0) and (2, 108) the 108 makes me think im completely wrong.
    Also, this question says show that d/dx = something, but what does d/dx mean?
    and if I had to sketch a graph of y=(x-1)^2, would it only cross the x axis at 1 and nowhere else?
    sorry for all the questions, haha
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    (Original post by jacksonmeg)
    Heya, could somebody check this question
    Find the coordinates of the maximum and minimum point on the curve y=x^2(5-x)^3, I got (5,0) and (2, 108) the 108 makes me think im completely wrong.
    Also, this question says show that d/dx = something, but what does d/dx mean?
    and if I had to sketch a graph of y=(x-1)^2, would it only cross the x axis at 1 and nowhere else?
    sorry for all the questions, haha
    d/dx means differentiate with respect to x what follows.

    d/dx of 2x is 2.

    Is that x to the power of 2(5-x), or x^2 multiplied by (5-x)?
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    (Original post by Liamnut)
    d/dx means differentiate with respect to x what follows.

    d/dx of 2x is 2.

    Is that x to the power of 2(5-x), or x^2 multiplied by (5-x)?
    x^2 multiplied by (5-x)^3
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    (Original post by jacksonmeg)
    Heya, could somebody check this question
    Find the coordinates of the maximum and minimum point on the curve y=x^2(5-x)^3, I got (5,0) and (2, 108) the 108 makes me think im completely wrong.
    Also, this question says show that d/dx = something, but what does d/dx mean?
    and if I had to sketch a graph of y=(x-1)^2, would it only cross the x axis at 1 and nowhere else?
    sorry for all the questions, haha
    If you mean y=x^2 (5-x)^3 then there are actually 3 stationary points, one of which is a point of inflection.

    Stationary points occur where there is 0 gradient i.e. \frac{dy}{dx} = 0

    Could you show us your working?
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    (Original post by Khallil)
    If you mean y=x^2 (5-x)^3 then there are actually 3 stationary points, one of which is a point of inflection.

    Stationary points occur where there is 0 gradient i.e. \frac{dy}{dx} = 0

    Could you show us your working?
    u= x^2 v = (5-x)^3

    du/dx = 2x dv/dx = -3(5-x)^2

    -3x^2(5-x)^2 + 2x(5-x)^3
    x(5-x)^2(-5x+10) = 0
    5-x = 0, x = 5, y = 0
    -5x+10 = 0, x = 2, y = 108
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    (Original post by jacksonmeg)
    u= x^2 v = (5-x)^3

    du/dx = 2x dv/dx = -3(5-x)^2

    -3x^2(5-x)^2 + 2x(5-x)^3
    x(5-x)^2(-5x+10) = 0
    5-x = 0, x = 5, y = 0
    -5x+10 = 0, x = 2, y = 108
    You seem to be forgetting x=0

    To verify that each of these points are maxima, minima or points of inflection, you need to evaluate the second derivative of the function with respect to x.
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    (Original post by Khallil)
    You seem to be forgetting x=0

    To verify that each of these points are maxima, minima or points of inflection, you need to evaluate the second derivative of the function with respect to x.
    oh yeah.... -idiot- thanks, the rest is ok though?
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    (Original post by jacksonmeg)
    oh yeah.... -idiot- thanks, the rest is ok though?
    yes
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    (Original post by jacksonmeg)
    oh yeah.... -idiot- thanks, the rest is ok though?
    Yep!
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    (Original post by Khallil)
    Yep!
    for the second differentiation, would u = x(5-x)^2 and v = -5x+10 ?
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    (Original post by Khallil)
    Yep!
    im getting 0, 0 and -90, are the 2 0s the turning points?
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    (Original post by jacksonmeg)
    for the second differentiation, would u = x(5-x)^2 and v = -5x+10 ?
    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.
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    (Original post by jacksonmeg)
    for the second differentiation, would u = x(5-x)^2 and v = -5x+10 ?
    Yep! You could even do the product rule with three terms, where f, g and h and functions of x:

    \dfrac{d}{dx} \left( fgh \right) = f'gh + fg'h + fgh'

    (Original post by jacksonmeg)
    im getting 0 and -90 so i cant of done it right
    If those are values of the second derivative, I got the same values. However, you're missing out the second derivative evaluated at x=0
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    (Original post by Khallil)
    Yep! You could even do the product rule with three terms, where f, g and h and functions of x:

    \dfrac{d}{dx} \left( fgh \right) = f'gh + fg'h + fgh'



    If those are values of the second derivative, I got the same values. However, you're missing out the second derivative evaluated at x=0
    yeah I just put 0 in and also got 0, so how would you differentiate between max and min? plot a graph?
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    (Original post by jacksonmeg)
    yeah I just put 0 in and also got 0, so how would you differentiate between max and min? plot a graph?
    What did you get for \dfrac{d^2y}{dx^2} \ ?
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    (Original post by Khallil)
    What did you get for \dfrac{d^2y}{dx^2} \ ?
    -5x(5-x)^2 -2x(5-x)(-5x+10)
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    (Original post by jacksonmeg)
    -5x(5-x)^2 -2x(5-x)(-5x+10)
    Try this as it may be simpler. Your first derivative is correct, so you probably made an error when trying to find the second derivative.

    \begin{aligned} \dfrac{dy}{dx} & = x(5-x)^2(-5x+10) \\ & = 5x(5-x)^2(2-x) \\ & = 5(5-x)^2(2x-x^2) \end{aligned}

    Now find \dfrac{d^2y}{dx^2} by letting u=(5-x)^2 and letting v=(2x-x^2)
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    (Original post by Khallil)
    Try this as it may be simpler. Your first derivative is correct, so you probably made an error when trying to find the second derivative.

    \begin{aligned} \dfrac{dy}{dx} & = x(5-x)^2(-5x+10) \\ & = 5x(5-x)^2(2-x) \\ & = 5(5-x)^2(2x-x^2) \end{aligned}

    Now find \dfrac{d^2y}{dx^2} by letting u=(5-x)^2 and letting v=(2x-x^2)
    5(5-x)^2(2-2x) - 10(5-x)(2x-x^2) ?
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    (Original post by jacksonmeg)
    5(5-x)^2(2-2x) - 10(5-x)(2x-x^2) ?
    Yep!

    To make everything neater, it usually helps to factorise out any terms that are common:

    \dfrac{d^2y}{dx^2} = 10(5-x)(2x^2 - 8x + 5)

    Have you heard of the second derivative test to determine the nature of stationary points?
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    (Original post by Khallil)
    Yep!

    To make everything neater, it usually helps to factorise out any terms that are common:

    \dfrac{d^2y}{dx^2} = 10(5-x)(2x^2 - 8x + 5)

    Have you heard of the second derivative test to determine the nature of stationary points?
    dont you put in the x values and the positive one is minimum, negative is max? I figured it didnt matter about simplifying since you just need to put in x values, and i usually go wrong anyway. lol
 
 
 

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