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    Need help with Q6

    I differentiated to get: (x^2)(3lnx+1)

    So x^2=0 and
    3lnx+1=0

    If I solve 3lnx+1=0 I get x= e^(-1/3) which I think is wrong

    So I don't know what to do?
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    Hello

    You need to recognise this is differentiation by parts. So,
    y=x^3lnx
    dy/dx=3x^2 (lnx) + x^3 (1/x)

    Therefore,
    dy/dx=3x^2 (lnx) + x^2
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    Then to find the min pt.
    gradient is 0
    3x^2 (lnx) + x^2 = 0

    x^2 ( 3x (lnx) + 1) = 0

    x=0 or 3x (lnx) = -1

    For you to have -1. 3x has to be negative as lnx cannot be negative. This means x has to be negative. But you can't ln a neg number. Hence that solution is NA.

    I'll explain in more detail if it didn't make sense

    EDIT: sorry I made a mistake
    it is 3lnx=-1

    so x=e^-1/3

    which means you are right! yipee and that solution is possible
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    (Original post by keerty)
    Then to find the min pt.
    gradient is 0
    3x^2 (lnx) + x^2 = 0

    x^2 ( 3x (lnx) + 1) = 0

    x=0 or 3x (lnx) = -1

    For you to have -1. 3x has to be negative as lnx cannot be negative. This means x has to be negative. But you can't ln a neg number. Hence that solution is NA.

    I'll explain in more detail if it didn't make sense
    Thats the part I'm stuck on

    So 3x(lnx) =-1 is NA

    So do u use x= 0? But the answer is -1/3e

    I'm confused
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    If you give me a moment I'll pm you the whole working
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    Hello,

    here it goes!

    so it's
    f(x) =x^3 (lnx)
    f'(x)= 3x^2 (lnx) + x^3 (1/x)
    f'(x)= 3x^2 (lnx) + x^2
    Let f'(x)=0

    x^2(3lnx +1)=0

    x=0 or lnx= -1/3
    therefore x=0 or x= e^-1/3
    If you got this answer you're right! If the book says otherwise you should check with your teacher. Though I'm pretty certain you're right.

    Rmb that after this you have to use the higher derivative to find out whether 0 or e^-1/3 is the min value
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    There is no way btw if you notice that you can get a neg value for x if you have lnx.
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    (Original post by Vorsah)
    Thats the part I'm stuck on

    So 3x(lnx) =-1 is NA

    So do u use x= 0? But the answer is -1/3e

    I'm confused
    The question asks for the minimum value of the function, NOT the value of x that produces that minimum!

    You need to stick your x value back into the function to check what the minimum value is
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    (Original post by keerty)
    Hello,

    If you got this answer you're right! If the book says otherwise you should check with your teacher. Though I'm pretty certain you're right.
    The OP needs to complete the question by finding the minimum value itself - see my previous update
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    Ah thanks for that didn't notice that myself.
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    (Original post by keerty)
    Hello,

    here it goes!

    so it's
    f(x) =x^3 (lnx)
    f'(x)= 3x^2 (lnx) + x^3 (1/x)
    f'(x)= 3x^2 (lnx) + x^2
    Let f'(x)=0

    x^2(3lnx +1)=0

    x=0 or lnx= -1/3
    therefore x=0 or x= e^-1/3
    If you got this answer you're right! If the book says otherwise you should check with your teacher. Though I'm pretty certain you're right.

    Rmb that after this you have to use the higher derivative to find out whether 0 or e^-1/3 is the min value
    My calculation tells me the same thing. After determining which of the x is a minimum value so you calculate the minimum value of f(x). Did you manage to get the answer, OP?

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    (Original post by sneha.vag)
    My calculation tells me the same thing. After determining which of the x is a minimum value so you calculate the minimum value of f(x). Did you manage to get the answer, OP?

    Posted from TSR Mobile
    Yeah
 
 
 
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