f(x)= (x^3)lnx f'(x) = (3x^2)(lnx) +(x^3)(1/x) = (3x^2)lnx + x^2 = x^2(3lnx+1) x^2 = 0 x=0 3lnx +1 =0 lnx =-1/3 x=e^(-1/3) is my work all correct? I was able to find find f''(x) of x^2(3lnx+1) so I don't know.
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I am trying to find out where f interval increase and decrease. so far I have INCREASE: ( __ , infinity) DECREASE: ( 0, __ )
Originally Posted by NeoSonata f(x)= (x^3)lnx f'(x) = (3x^2)(lnx) +(x^3)(1/x) = (3x^2)lnx + x^2 = x^2(3lnx+1) x^2 = 0 x=0 3lnx +1 =0 lnx =-1/3 x=e^(-1/3) is my work all correct? Yeh, this looks good, now graph the function using technology, is $\displaystyle \displystyle \frac{1}{\sqrt[3]{e}}\approx 0.72$ in the neighbourhood of your turning point?
yeah.. ___________________ 0 ---- .72 ++++++ so it should be increasing from (.72, infinity) and decreasing from (0, .72) ? web-assign will not accept so probably the system is incorrect.
Originally Posted by NeoSonata so it should be increasing from (.72, infinity) and decreasing from (0, .72) ? That's how I see it.
Maybe WebAssign needs more digits for accuracy, or maybe it needs the exact result.
It might do, I'm not aware of this software. If you were doing a written exam the exact value of e^(-1/3) will always be best.
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