C3 help needed

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

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.

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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