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

1. ## 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..

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

Originally Posted by NeoSonata
f increase at (e^(-1/3), infinity)
Right.

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

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

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 $x=e^{-\frac{5}{6}}$ wherefore $f''(x)=0$. So what's the inflection point (x and y coordinate)? ...
You can now determine where the functions is convex or concave.

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

we have to check f '
f ' is negative if $\ln{x}$ < $\frac{1}{3}$
gives x < $e^{\frac{1}{3}$
therefore f is decreasing if x $\in$ ( 0 , $e^{\frac{1}{3}$ )
and f is increasing if x $\in$ ( $e^{\frac{1}{3}$ , $\infty$)