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anonymous
 one year ago
Use a graphing calculator to approximate all of the functions real zeros. Round to 4 decimal places.
f(x)= 3x6−5x5−4x3+x2+x+1
anonymous
 one year ago
Use a graphing calculator to approximate all of the functions real zeros. Round to 4 decimal places. f(x)= 3x6−5x5−4x3+x2+x+1

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mathmate
 one year ago
Best ResponseYou've already chosen the best response.0@_alex_urena_ Have you graphed the function yet? How many zeroes do you see?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah i graphed it but what do you mean by zeros?

mathmate
 one year ago
Best ResponseYou've already chosen the best response.0A zero in a polynomial (or any other) equation is where the graph intersects the xaxis. In general, a polynomial of 6th degree (your case) has a maximum of 6 zeroes.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah one of them is in between 0 and 1 and the other is on 2

mathmate
 one year ago
Best ResponseYou've already chosen the best response.0Good, so there are two (real) zeroes.

mathmate
 one year ago
Best ResponseYou've already chosen the best response.0Have you learned Newton's method, or the bisection method, or anything similar to those?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0uhm i dont think so. never heard those names

mathmate
 one year ago
Best ResponseYou've already chosen the best response.0Can you do trial and error?

mathmate
 one year ago
Best ResponseYou've already chosen the best response.0You're trying to solve for the zeroes of f(x)=3x6−5x5−4x3+x2+x+1 which means that where f(x) crosses the x axis. You know that there's one between 0 and 1. so you can try with f(0)=1, f(1)=3. One is positive and the other is negative right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah but idk where you got three from

mathmate
 one year ago
Best ResponseYou've already chosen the best response.0f(1) means put x=1 in f(x), or f(1)=3(1)^6−5(1)^5−4(1)^3+(1)^2+(1)+1=3

mathmate
 one year ago
Best ResponseYou've already chosen the best response.0That means the curve has to cross the xaxis (y=0) somewhere between 0 and 1.

mathmate
 one year ago
Best ResponseYou've already chosen the best response.0So you would try f(1/2), (between 0 and 1) to see if it is positive or negative. Use your calculator to find f(1/2)=1.14. So the solution f(x)=0 is between 1/2 and 1 because f(1/2)>0, and f(1)<0.

mathmate
 one year ago
Best ResponseYou've already chosen the best response.0So we have narrowed down the gap by a factor of two. If you now repeat the same "trial" and "error", you can narrow down the zero to a much more accurate value.

mathmate
 one year ago
Best ResponseYou've already chosen the best response.0Your next try should be (0.5+1)/2=0.75. So try f(0.75) and see what you have. The key is to narrow down the gap. If f(0.75) is positive, your next try is x=(0.75+1)/2=0.875, and if f(0.75) is negative, your next try is x=(0.5+0.75)/2=0.625.
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