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anonymous
 one year ago
Use synthetic division to determine whether the number k is an upper or lower bound (as specified) for the real zeros of the function f.
k = 4; f(x) = 2x^3  2x^2  3x  5; Lower bound?
anonymous
 one year ago
Use synthetic division to determine whether the number k is an upper or lower bound (as specified) for the real zeros of the function f. k = 4; f(x) = 2x^3  2x^2  3x  5; Lower bound?

This Question is Closed

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes or no are the answer choices

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1So, you have to solve for the zeros of the function. What are they?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm honestly not sure how to find them

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1use demos graph to see where they cut xaxis

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So would the answer be no since it's (2.2,0) and 4 would be a higher bound and not lower bound?

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1we don't have 0. please check

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0what do you mean? it cut the xaxis at 2.2

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1but you can use it. now find f(0) =?

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1replace 0 into f(x) , what do you get?

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1ok, so f(0) =5 < 0 , right? now f (2) =?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So what does that mean

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1that mean from 0 to 2 you still not pass the xaxis, since f(x) <0 for both. Now test f(3)

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1yes, so from 2 to 3 , f(x) go from negative to positive, that is you have one zero on that interval, right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Again, the lesson didnt really explain this stuff so I'm not sure how to use that

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1you don't know synthetic?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0not completely, looking it up now trying to learn real quick

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1dw:1442537779766:dw got it?

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1the last one, you do, the same

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0would it be 4*21 = 84 or 4*24 =96?

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1you take 4 * the bottom row, put the result on the middle row of the new column, then add them up to get the new number.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0that's easy then haha

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1ok, now, 4 >0 , right?

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1look at the bottom row, they are 2,6,21,79 right?

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1and all them are positive, right?

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1Hence 4 is upper bound for the zeros

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0woah i see, a lot goes into it

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so back to the original question the answer would be no

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok thanks for all of the help, what would happen if one of them was negative?

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1you test, if k <0, the bottom line is +, ,+, alternatively, then k is lower bound

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1learn from it http://www.howetwo.com/mathematicat/bound.html
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