Need help with a 5th order polynomial exercise

[sun303]

Can anyone help? I have no idea how to tackle this exercise…

 

[topsquark]

It's symmetric so it's either odd or even. If it were even then the [imath]x^5[/imath] term would spoil that, so it must be odd. Thus all the even power terms are gone and you are looking at [imath]ax^5 + bx^3 + cx[/imath]. Can you continue?

-Dan

 

[Dr.Peterson]

View attachment 31245
Can anyone help? I have no idea how to tackle this exercise…

As topsquark says, "symmetrical with respect to the origin" means it's odd; "touches the x-axis" tells you both y and y' at the indicated point(s). So you'll have three equations with three unknown parameters.

This can also be solved without calculus, using the fact that symmetry gives you a total of five points on the curve, and touching the axis provides the factored form. I find this approach considerably easier.

 

[AvgStudent]

It's symmetric so it's either odd or even. If it were even then the [imath]x^5[/imath] term would spoil that, so it must be odd. Thus all the even power terms are gone and you are looking at [imath]ax^5 + bx^3 + cx[/imath]. Can you continue?

-Dan

Why don't we consider a constant? [imath]ax^5+bx^3+cx+d[/imath]. Is it because [imath]d=dx^0[/imath] and [imath]x^0[/imath] is considered as an even powered term?

 

[Deleted member 4993]

Why don't we consider a constant? [imath]ax^5+bx^3+cx+d[/imath]. Is it because [imath]d=dx^0[/imath] and [imath]x^0[/imath] is considered as an even powered term?

Do you think the constant term could be symmetric about the origin?

 

[AvgStudent]

Do you think the constant term could be symmetric about the origin?

Good point?. If [imath]d\neq 0[/imath], then [imath]f(0)=d[/imath]. It's no longer at the origin, but at [imath](0,d)[/imath] instead, because we shifted the function up/down. Thanks.