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

= 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 &minus; 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 $$p\ge0$$, as it follows looking at its first derivative.pm a  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)


 * Whether injective or not, most (almost all?) cubics cannot be put in the form f (x) = m (x &minus; 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:


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


 * 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)