f(x) = (x^3)(lnx)

f'(x) = (3x^2)(lnx) + (x^3)(1/x)

how do solve for x when

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

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- Jul 25th 2011, 02:57 AMNeoSonataneed help taking dervative of (x^3)(lnx) and solving for x
f(x) = (x^3)(lnx)

f'(x) = (3x^2)(lnx) + (x^3)(1/x)

how do solve for x when

(3x^2)(lnx) + (x^3)(1/x) = 0 - Jul 25th 2011, 03:04 AMSironRe: need help taking dervative of (x^3)(lnx) and solving for x
You have:

$\displaystyle f'(x)=3x^2\cdot \ln(x)+x^2=0$

It can be very useful to wright:

$\displaystyle x^2[3\ln(x)+1]=0$

EDIT:

Also remember: $\displaystyle \ln(0)$ is undefined! - Jul 25th 2011, 03:15 AMNeoSonataRe: need help taking dervative of (x^3)(lnx) and solving for x
so x = 0 and:

lnx = -1/3

x = e^(-1/3)

correct?

Thanks. - Jul 25th 2011, 03:18 AMSironRe: need help taking dervative of (x^3)(lnx) and solving for x
Almost, $\displaystyle x=e^{\frac{-1}{3}}$ is a solution, but see my EDIT, $\displaystyle \ln(0)$ is undefined, so $\displaystyle x=0$ is not a solution.

What you always have to do if you solve logarithmic equations is first determine where the function is defined and undefined (the 'existence conditions'), so at the end you can compare with your 'existence conditions' if the solution(s) satisfies.

(Sorry, I had to say this before :))