Reconstructing the parameters of 2ax³+bx²+cx

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Let $f(x) = 2ax^3 + bx^2 + cx$.

• (a) find $a,b,c$ for an inflection at $P(0,1)$ and a relative extremum at $x=1$ (state whether max or min);
• (b) find $a,b,c$ for a relative extremum at $P(3,0)$ and one at $x=-1$;
• (c) find $a,b,c$ for an inflection at $P(-1,2)$ with tangent of slope $2$.

Solution

(a) Inconsistent conditions. Note immediately that $f(0) = 0$ for every choice of $a,b,c$: the graph always passes through the origin, so an inflection at $P(0,1)$ (ordinate $1$) is impossible — it contradicts the assigned ordinate. The only consistent geometric conditions give: inflection at $x=0$ → $f''(0) = 2b = 0 \Rightarrow b = 0$; stationary point at $x=1$ → $f'(1) = 6a + c = 0 \Rightarrow c = -6a$. This leaves the family $f(x) = 2a\,(x^3 - 3x)$ with $a$ free. Note: the ordinate constraint on the inflection is incompatible.

(b) Underdetermined system. With only two conditions ($f'(3)=0$ and $f'(-1)=0$) and three unknowns, the system does not fix $(a,b,c)$. Also imposing $f(3)=0$ (the point $P(3,0)$ lies on the graph) together with $f'(3)=0$ and $f'(-1)=0$ yields only the trivial solution $a=0$ (which degenerates the cubic). The exercise is ill-posed: the conditions do not single out a proper cubic.

(c) Solvable case. Inflection at $x=-1$: $f''(-1) = 12a\cdot(-1) + 2b = -12a + 2b = 0 \Rightarrow b = 6a$. Passing through $(-1,2)$: $f(-1) = -2a + b - c = 2$. Slope $2$ at the inflection: $f'(-1) = 6a - 2b + c = 2$. Substituting $b = 6a$: $-2a + 6a - c = 2 \Rightarrow 4a - c = 2, \qquad 6a - 12a + c = 2 \Rightarrow -6a + c = 2.$ Adding: $-2a = 4 \Rightarrow a = -2$, hence $b = -12$ and $c = 4a - 2 = -10$. $\boxed{a = -2,\quad b = -12,\quad c = -10}$