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

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The discussion focuses on determining the concavity and inflection points of the polynomial function y=3x^5 - 5x^4 - 60x^3 + 80. The second derivative, y''=60x(x+2)(x-3), reveals three roots at x=-2, x=0, and x=3, indicating potential inflection points. The sign of the second derivative alternates across the intervals created by these roots, leading to two intervals of concavity up and two of concavity down. Specifically, the function is concave down on (-∞, -2) and (0, 3), and concave up on (-2, 0) and (3, ∞). The analysis confirms the identification of inflection points and the behavior of the function across its domain.
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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 $y$
 
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Factoring the second derivative (which you correctly computed), we find:

$$y''(x)=60x(x+2)(x-3)$$

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

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

Given that:

$$y''(x)=60x(x+2)(x-3)$$

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):

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

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

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

$$(-\infty,-2)\implies -$$
 
sorry just thought the table was cute

but your input was very helpful