Question

Let f:x tends to amx^m+am-1x^m-1+··· +a1x+a0 be a polynomial of even degree. If amao < 0 show that the equation f(x) = 0 has at least two real roots.

Answer 1

could you please format the question again using latex ?

Answer 2

question 3 please

Answer 3

Just wanted to check if u are still online ?

Answer 4

@freckles

Answer 5

im not getting ideas... this tagging has stopped worked again..

Answer 6

Can u try something out of it, i have a quiz n it tomorrow. Thanks.

Answer 7

Give an example of a nonconstant continuous function for which Intermediate value holds and such that there are infinitely many values x0 where f(x0) = c.

Answer 8

can u get me this one ?

Answer 9

@Kainui

Answer 10

Anybody please.

Answer 11

Well for any odd polynomial you have to have at least 1 real root because you will have something that for negative values is negative and for positive values is positive.

Similarly for even polynomials if you know that the constant term is less than 0 there has to be 2 roots since all even polynomials are positive for negative and positive terms.

The simplest example of this would be something like

f(x)=x^2-1

See how at f(0)=-1 and we know since it's even the rest of the terms are positive outside of that. Now if we look at the leading coefficient, this is the dominating term so it will ultimately decide if the end behavior is positive or negative. So we can see

g(x)=-2x^2+3

will end pointing down so the +1 will allow us to be higher. This is just the exact opposite of the last situation I made. Here are some pictures of these two functions: