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Babynini
 one year ago
Prove the equation has at least 1 real root.
Babynini
 one year ago
Prove the equation has at least 1 real root.

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ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1Have you tried using IVT ?

Babynini
 one year ago
Best ResponseYou've already chosen the best response.0Noo. I'm not sure where to start.

Babynini
 one year ago
Best ResponseYou've already chosen the best response.0I don't know what IVT is o.o

dan815
 one year ago
Best ResponseYou've already chosen the best response.3find that the function has a positive and a negative value

dan815
 one year ago
Best ResponseYou've already chosen the best response.3or use the fact that it's an odd functions, so the ends, have one going to infnite and the other going to negative infinity

dan815
 one year ago
Best ResponseYou've already chosen the best response.3and thsi is an analytical function meaning there are no holes, and its connected so it must have passed through 0 to do this

Babynini
 one year ago
Best ResponseYou've already chosen the best response.0Ahh k. ^ makes sense thus far.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1If you're not familiar with IVT yet, you may also simply use the fundamental theorem of algebra : A polynomial of degree "n" has exactly "n" roots and out of these "n" roots, the complex roots always come in conjugate pairs.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1since the polynomial in question has a degree of 5, which is odd, it follows that at least one root must be real.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1Look at the equayion in problem #56 first convinvce yourself that it is a polynomial equation

dan815
 one year ago
Best ResponseYou've already chosen the best response.3just do this f(10000) = positive f(1000000) = negative you are done

dan815
 one year ago
Best ResponseYou've already chosen the best response.3since its a continuos function these 2 points are connected in between in some way, and the line must obviously pass y=0

dan815
 one year ago
Best ResponseYou've already chosen the best response.3the curve* must pass if u are picky .

Babynini
 one year ago
Best ResponseYou've already chosen the best response.0Haha k. Well, what does it mean for it to have a real root? that just means it passes through 0?

dan815
 one year ago
Best ResponseYou've already chosen the best response.3a real root means u have an x value where that intersects the x axis

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1real roots are same as x intercepts in the graph

Babynini
 one year ago
Best ResponseYou've already chosen the best response.0Ahhh k. So as soon as I recognize it's a polynomial then I basically know it has root? . why do I want to go about proving it then?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1for example, below graph has 3 x intercepts, so we say the number of real roots are 3 dw:1444528871031:dw

dan815
 one year ago
Best ResponseYou've already chosen the best response.3highest is odd degree* polynomial

dan815
 one year ago
Best ResponseYou've already chosen the best response.3even highest degree polynomials dont need to have real root

Babynini
 one year ago
Best ResponseYou've already chosen the best response.0I see, I see. Ok, this is making much more sense.

Babynini
 one year ago
Best ResponseYou've already chosen the best response.0So for part a, I legit just have to say: "since the polynomial in question has a degree of 5, which is odd, it follows that at least one root must be real."

dan815
 one year ago
Best ResponseYou've already chosen the best response.3you might need more support, the better way is to show that htere is a positive and a negative value

dan815
 one year ago
Best ResponseYou've already chosen the best response.3and say since its continuous.. it must pass thru xaxis

dan815
 one year ago
Best ResponseYou've already chosen the best response.3you cant just be saying oh its degree 5, so it just works, u are going to have to link polynomial functiosn to their odd function counter parts

dan815
 one year ago
Best ResponseYou've already chosen the best response.3you are going to have to state something like since its a composition of odd and even functions x^5, ax^4, bx^3,cx^3,dx,3 as x goes to infinite the additions of the other degrees will pare in comparision and the end points will exhibit the odd degree characteristics

Babynini
 one year ago
Best ResponseYou've already chosen the best response.0hm ok. Should I do any equations to prove it?

dan815
 one year ago
Best ResponseYou've already chosen the best response.3u could do x^5 + ax^4 + bx^3 + cx^3 + dx +3 =0 divide everything by x^4 x+a+b/x + c/x^2 + d/x^3 + 3/x^4=0 now u can check x=inf and inf if u set x=inf ull end up with inf +a if u set x=inf ull end up with inf +a

dan815
 one year ago
Best ResponseYou've already chosen the best response.3so u have shown that this functions goes to inf and inf at its end points

Babynini
 one year ago
Best ResponseYou've already chosen the best response.0hmm ok. Why did you add all those extra numbers? o.o

Babynini
 one year ago
Best ResponseYou've already chosen the best response.0Yeah. If the original is: x^5x^2+2x+3

dan815
 one year ago
Best ResponseYou've already chosen the best response.3its just more general, u can plug in the values for a b c d respectively

dan815
 one year ago
Best ResponseYou've already chosen the best response.3i couldnt remember your question exactly so

Babynini
 one year ago
Best ResponseYou've already chosen the best response.0^ thanks so much. I think I get it :) looking over notes and looking at all you said.
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