How many turning points do you see in this graph? Graph link is in the first "slot" below this question. Thanks.
http://www.wolframalpha.com/input/?i=y%3D-%28x+%E2%80%93+1%29%28x+%2B+2%29%5E3
I see two - one at x=-2, one at roughly x=-0.5.
Do you mean inflection points?
@ opiesche --> Does your answer mesh with this theory (at the link) ----------------------------------- Maximum Number of Turning Points. The maximum number of turning points is one less than the degree of the polynomial. So, if the degree is n, the maximum number of turning points is n–1. Minimum Number of Turning Points. The minimum number of turning points depends on whether the degree, n > 0, is odd or even. If n is even, there is at least one turning point but if n is odd, there may not be any. http://webgraphing.com/polynomialtricksoftrade.jsp
@ Fool ---> I don't know because the question came from another asker. Here is the question as originally stated ---> http://openstudy.com/users/directrix#/updates/4f376bdee4b0fc0c1a0d4e96
Let's check :P http://www.wolframalpha.com/input/?i=d%2Fdx++-%28x+%E2%80%93+1%29%28x+%2B+2%29^3 The derivative changes signs at -2 and ~+0.25, not -0.5. Oh how the eyes can deceive ;)
@Directrix: I am not sure but I think the inflection point is the only way out for a precise answer.
Differentiate the function to and find real roots of the derivative All values are the inflection points
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2 turning points
Correction 1 inflextion 1 maxima
At -2, there's a double and then a single at 1/4 --> zeros of y'. I look at the graph and think I see 1 inflection point and 1 max. Have you studied "terrace points?" That was mentioned in class.
Might have but knowing by another name
I'm assuming that's where d' becomes 0 and stays there?
y' of course. It's late.
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