Testwiki:Reference desk/Archives/Mathematics/2019 January 17

{| width = "100%"

|- ! colspan="3" align="center" | Mathematics desk |- ! width="20%" align="left" | < January 16 ! width="25%" align="center"|<< Dec | January | Feb >> ! width="20%" align="right" |Current desk > |}

Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is a transcluded archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

January 17

Injective cubic polynomial functions

Is every cubic polynomial f (x) with real coefficients that is injective as a map from the real numbers to itself of the form f (x) = a (x − h)³ + k for some real numbers a, h, and k with a nonzero? GeoffreyT2000 (talk) 02:33, 17 January 2019 (UTC)

No because your polynomial's second derivative has a zero in $x=h$, which is not the case, e.g. of $x^3+x$, still injective as a map from the real line to itself. Say that up to a translation $u=x-h$ any cubic polynomial is $a[u^3+pu+q]$ with $a$ nonzero; it is injective if and only if $p\ge 0$, as it follows looking at its first derivative.pma 08:03, 17 January 2019 (UTC)

Alternatively, f(x) = ax³+bx²+cx+d, a≠0, is injective iff b²≤3ac. --RDBury (talk) 10:59, 17 January 2019 (UTC)

Template:Ping Whether injective or not, most (almost all?) cubics cannot be put in the form f (x) = m (x − h)³ + k. Let the cubic be f(x) = ax³+bx²+cx+d, a≠0. Equate this to the expanded form of the required form:

$m(x^3-3h{x}^2+3h^{2}x-h^3)+k$ ≡ $a{x}^3+b{x}^2+cx+d.$

For this to hold as an identity (hence the ≡ sign), the coefficients of like terms must always be equal (as per the article Equating coefficients). Thus we have four parameter conditions to be met, in only the three unknown parameters $h,k,m.$ Loraof (talk) 17:28, 22 January 2019 (UTC)

• The parameter condition on the coefficents of the given polynomial for these four equations in three unknowns to be consistent is 3ac=b². As RDBury noted above, this case is injective. Loraof (talk) 18:32, 22 January 2019 (UTC)