# Cubic Functions

• Mar 9th 2013, 09:00 AM
HelenMc9
Cubic Functions
A cubic function f is defined for x ∈ R as f (x) = x3+ (1 - k2​)x + k, where k is a constant. Show that -k is a root of f.

How would I go about doing this? Help appreciated! :)
• Mar 9th 2013, 09:17 AM
Plato
Re: Cubic Functions
Quote:

Originally Posted by HelenMc9
A cubic function f is defined for x ∈ R as f (x) = x3+ (1 - k2​)x + k, where k is a constant. Show that -k is a root of f.

How would I go about doing this? Help appreciated!

What is \$\displaystyle f(-k)=~?\$
• Mar 9th 2013, 09:24 AM
HelenMc9
Re: Cubic Functions
Aah yes it's equal to zero because -k is a root... I know what to do now! :) thanks!!
• Mar 9th 2013, 10:17 AM
HallsofIvy
Re: Cubic Functions
By the way, "root" is the wrong word here. You can find a "root" of an equation such as f(x)= 0, or a "zero" of a function, that being, of course, the root of the equation f(x)= 0.
• Mar 9th 2013, 10:42 AM
Plato
Re: Cubic Functions
Quote:

Originally Posted by HallsofIvy
By the way, "root" is the wrong word here. .

There does seem to be general agreement on that.
See zero here and root there.