It is confusing. You never mentioned what 'y' is? Y=f(x). And since the graph of y=f(x) touches the x-axis at least once y=0 at that point, you concluded that there exists one real root. Throughout the graph of the plane 'x' or the input values were all real. how did you say that? Please explain if you can
just makes sure understand this: beacuse of the end behavior of odd functions specifically that will go either from quad 3 to 1 or 2 to 4 they will have to cross the x axis and therefore have at least 1 root ( since root is when the point at the x axis. where as even functions end behavior goes either from quad 3 to 4 or 1 2 to 1 the graph this way does not cross the x axis and therefore it is possible for even degree polynomials to have no roots. did i get this right? did miss anything important here?
You explained this beautifully, thank you!
Well explained sir 🙏.
Thank you!!!
It is confusing. You never mentioned what 'y' is? Y=f(x). And since the graph of y=f(x) touches the x-axis at least once y=0 at that point, you concluded that there exists one real root. Throughout the graph of the plane 'x' or the input values were all real. how did you say that? Please explain if you can
just makes sure understand this: beacuse of the end behavior of odd functions specifically that will go either from quad 3 to 1 or 2 to 4 they will have to cross the x axis and therefore have at least 1 root ( since root is when the point at the x axis. where as even functions end behavior goes either from quad 3 to 4 or 1 2 to 1 the graph this way does not cross the x axis and therefore it is possible for even degree polynomials to have no roots.
did i get this right? did miss anything important here?
Thank you so much!
Are we solve this question with intermediate value theorem?
Same can be done for even number
Even degree polynomials may not have any real root.
x^2 + 1