At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
Have you tried using IVT ?
Noo. I'm not sure where to start.
I don't know what IVT is o.o
find that the function has a positive and a negative value
or use the fact that it's an odd functions, so the ends, have one going to infnite and the other going to negative infinity
and thsi is an analytical function meaning there are no holes, and its connected so it must have passed through 0 to do this
Ahh k. ^ makes sense thus far.
If 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.
since the polynomial in question has a degree of 5, which is odd, it follows that at least one root must be real.
degree of 5? hm?
Look at the equayion in problem #56 first convinvce yourself that it is a polynomial equation
just do this f(10000) = positive f(-1000000) = negative you are done
since its a continuos function these 2 points are connected in between in some way, and the line must obviously pass y=0
the curve* must pass if u are picky -.-
Haha k. Well, what does it mean for it to have a real root? that just means it passes through 0?
thats what a root is
a real root means u have an x value where that intersects the x axis
real roots are same as x intercepts in the graph
Ahhh 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?
for example, below graph has 3 x intercepts, so we say the number of real roots are 3 |dw:1444528871031:dw|
highest is odd degree* polynomial
even highest degree polynomials dont need to have real root
I see, I see. Ok, this is making much more sense.
So 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."
you might need more support, the better way is to show that htere is a positive and a negative value
and say since its continuous.. it must pass thru xaxis
you 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
you 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
hm ok. Should I do any equations to prove it?
u 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
so u have shown that this functions goes to inf and -inf at its end points
hmm ok. Why did you add all those extra numbers? o.o
Yeah. If the original is: x^5-x^2+2x+3
its just more general, u can plug in the values for a b c d respectively
i couldnt remember your question exactly so
^ thanks so much. I think I get it :) looking over notes and looking at all you said.