Find Where f Increase, Decrease, Local Minimum, Inflection Point, concave up and down

f(x) = x^3 lnx

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

x= e^(-1/3)

left side of e^(-1/3) is negative, right side is positive so

**f increase at (e^(-1/3), infinity)**

f decrease (-infinity , e^(-1/3)) correct?

Local Minimum:

f(e^(-1/3)) = .123 so local minimum is:

**.123 ** correct?

**Inflection point:**

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

f "(X) = 6xlnx + 3x + 2x

= x (6lnx + 5) = 0

x= e^(-5/6)

----- e^(-5/6) ++++

I don't know what to do from here..

Re: Find Where f Increase, Decrease, Local Minimum, Inflection Point, concave up and

Quote:

Originally Posted by

**NeoSonata** f increase at (e^(-1/3), infinity)

Right.

Quote:

f decrease (-infinity , e^(-1/3)) correct?

Wrong, $\displaystyle f$ is defined in $\displaystyle (0,+\infty)$. As a consequence, it is decreasing in $\displaystyle (0,e^{-1/3})$ .

Re: Find Where f Increase, Decrease, Local Minimum, Inflection Point, concave up and

Quote:

Originally Posted by

**NeoSonata** **Inflection point:**

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

f "(X) = 6xlnx + 3x + 2x

= x (6lnx + 5) = 0

x= e^(-5/6)

----- e^(-5/6) ++++

I don't know what to do from here..

You have found $\displaystyle x=e^{-\frac{5}{6}}$ wherefore $\displaystyle f''(x)=0$. So what's the inflection point (x and y coordinate)? ...

You can now determine where the functions is convex or concave.

Re: Find Where f Increase, Decrease, Local Minimum, Inflection Point, concave up and

we have to check f '

f ' is negative if $\displaystyle \ln{x}$ < $\displaystyle \frac{1}{3}$

gives x < $\displaystyle e^{\frac{1}{3}$

therefore f is decreasing if x $\displaystyle \in$ ( 0 , $\displaystyle e^{\frac{1}{3}$ )

and f is increasing if x $\displaystyle \in$ ( $\displaystyle e^{\frac{1}{3}$ , $\displaystyle \infty$)