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

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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 $a{\big [}u^{3}+pu+q{\big ]}$ with $a$ nonzero; it is injective if and only if $p\geq 0$, 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)

@GeoffreyT2000: 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:

$m(x^{3}-3hx^{2}+3h^{2}x-h^{3})+k$ ≡ $ax^{3}+bx^{2}+cx+d.$

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)