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Mathematics 18 Online
OpenStudy (anonymous):

Tell me whether or not the following are upper or lower bounds of the function F(x) = 2x^3 + 4x^2 - 2x - 4 x= -1 x=21 So I synthetically divided both, but x=-1 is a zero of the function, so how can I tell whether it's an upper or lower bound?

OpenStudy (amoodarya):

what is the limit of x?

OpenStudy (anonymous):

No idea

OpenStudy (amoodarya):

if x is not bounded f is not bounded yet (F(x) = 2x^3 + 4x^2 - 2x - 4) I mean that function

OpenStudy (anonymous):

Sorry I'm not exactly sure what you're saying

OpenStudy (amoodarya):

suppose f(x)=x if x is not limited in (a,b) or [a,b] there is not a bound!

OpenStudy (anonymous):

Oh I understand what you're saying now. Would it be possible for -1 and 20 to be the bounds? Because what I posted is all that's in the question. If I understand correctly, a line of alternating positive and negative numbers is a lower bound, and all positives is an upper bound, but when I synthetically divide -1. it is + + - +, and the remainder is also zero, indicating that -1 is also a zero of the polynomial.

OpenStudy (amoodarya):

if we suppose that x belongs to {-1,21] so check f(-1) f(21) f(root of f') f(1/3) f(-1) dulicate f'=6x^2+8x-2=0 so x=-1 ,+2/6

OpenStudy (tkhunny):

This is a question concerning zeros of the function, not the limits of the function. Upper Bounds of zeros have ALL positive values across the bottom of the Synthetic Division Tableau. Lower Bounds of zeros have strictly ALTERNATING values across the bottom of the Synthetic Division Tableau. Alternatively, substitute x = -x and search for the upper bounds of the new function. It is presumed that Upper Bounds of zeros are positive and Lower Bounds of zeros are negative. The general idea is that zeros of polynomials tend to happen in the neighborhood of x = 0. The question is how far from x = 0 should we look?

OpenStudy (anonymous):

I think I got it guys, the answer is simply that -1 is not an upper or lower bound of the function. Thanks for the help though :)

OpenStudy (tkhunny):

The third value in the synthetic division tableau is negative. -1 is not an upper bound. Good work.

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