1. ## Locating Concavity

$
\displaystyle f(x) = x^{4}-3x^{3}+3x^{2}
$

Find the open intervals for which concavity is open upwards.

$
\displaystyle f'(x)=4x^3-9x^2+6x
$

$
\displaystyle f''(x)=12x^2-18x-6
$

I set f''(x)=0
$

\displaystyle 6(2x^2-3x-1)=0
$

$
\displaystyle \frac{3\pm\sqrt{17}}{4}
$

So when I test the points to the left & to the right, and in-between, the positive intervals are:
$
\displaystyle (-\infty,\frac{3-\sqrt{17}}{4}],[\frac{3+\sqrt{17}}{4},\infty)
$

which is apparently wrong. Where did I mess up? Thanks!

2. Well, one mistake is in taking the second derivative: the derivative of +6x is not -6.

3. Concave up if the second derivative is positive, concave down if the second derivative is negative. If the second derivative is zero, your usually (but not always) at a point where the concavity is changing from up to down or from down to up. (An example of the not always case is $y=x^{4}$ at the origin.)

4. Got it. Thanks.

However, this one is question has got me stumped.

$
f(x) = -x^{4}\ln\!\left(x\right)
$

I need to find the intervals where f is increasing and decreasing.

So I start by taking the first derivative:
$
f'(x)=-3x^3lnx-x^3
$

$
-x^3(3lnx+1)=0
$

$
x=0 , e^{\frac{-1}{3}}
$

since 0 is not in the natural domain, it is not a critical point right?

When I plug in points on a number line, it shows that my function is increasing on:
$(0,e^{\frac{-1}{3}}]$ and decreasing on $[e^{\frac{-1}{3}},\infty)$

This is wrong. Where did I mess up?

Thanks!

5. You made a mistake with the differentiation again !