Concave Up or Down? Identifying Inflection Points in a Polynomial Function

[karush]

Determine the intervals on which $$y=3x^5 - 5x^4 - 60x^3 + 80$$
$$y'=15{x}^{4}-20{x}^{3}-180{x}^{2}$$
$$y''=60{x}^{3}-60{x}^{2}-360x$$
$$y''\left(0\right)=-2,0,3$$

Is concave up or concave down, identity inflection points. Mutiple choiceA. The function is concave up on___ and concave down on____B. The function is concave up on__C. The function is concave down on___

From observation don't see any concave up, but there is a slope of zero on $y$

 

[MarkFL]

Factoring the second derivative (which you correctly computed), we find:

\(\displaystyle y''(x)=60x(x+2)(x-3)\)

And so you have also correctly identified the roots of the second derivative. Now, because each of these roots is of odd multiplicity, we know the sign of $y''$ will alternate across all 4 intervals the 3 roots make in the domain of $y$, and so we know there will be an inflection point at each root. (Why do we know this?)

So, what you want to do is compute the sign of $y''$ within anyone of the intervals, and then fill in the rest, knowing they must alternate. Then you can state that $y$ is concave up in the positive intervals and concave down in the negative intervals. We know then that there will be 2 intervals on which $y$ is concave up and 2 intervals on which $y$ is concave down.

Can you proceed?

 

[karush]

View attachment 5304
So from the intervals of
$$\left(-\infty, - 2\right)$$ the sign is -
$$\left(-2 , 0\right)$$ the sign is +
$$\left(0 , 3\right)$$ the sign is -
$$\left(3 , \infty\right)$$ the sign is +
I hope

Is there any to change line types on Desmos, colors still are hard to distinguish

 

[MarkFL]

I wouldn't turn to technology for this...we have the intervals:

\(\displaystyle (-\infty,-2),\,(-2,0),\,(0,3),\,(3,\infty)\)

Given that:

\(\displaystyle y''(x)=60x(x+2)(x-3)\)

If we check the rightmost interval with an $x$ value in that interval (say $x=4$), then the signs of the factors will be:

(+)(+)(+) = +

So, we know the sign in the rightmost interval is positive, thus we must have (letting them alternate):

\(\displaystyle (3,\infty)\implies +\)

\(\displaystyle (0,3)\implies -\)

\(\displaystyle (-2,0)\implies +\)

\(\displaystyle (-\infty,-2)\implies -\)

 

[karush]

sorry just thought the table was cute

but your input was very helpful