Cubic ax³+bx²+cx: study and reconstruction from constraints

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

• (a) with $a=-1,\ b=13,\ c=-40$: study the sign and the intersections with the axes;
• (b) find relative maxima/minima, the inflection point and the tangent line at the inflection;
• (c) sketch the graph;
• (d) find $a,b,c$ so that the graph has an inflection at $P(1,2)$ and a relative extremum at $x=0$ (state whether max or min);
• (e) find $a,b,c$ for a relative maximum at $P(-3,3)$ and a relative minimum at $x=0$;
• (f) find $a,b,c$ for an inflection at $P(1,1)$ with horizontal tangent.

Solution

(a) Sign and zeros. $f(x) = -x^3 + 13x^2 - 40x = -x(x-5)(x-8)$; zeros $x=0,\ 5,\ 8$.

(b) Extrema, inflection, tangent. $f'(x) = -3x^2 + 26x - 40 = 0 \Rightarrow x = 2,\ x = \tfrac{20}{3}$. At $x=2$: minimum, $f(2) = -8 + 52 - 80 = -36$. At $x=\tfrac{20}{3}$: maximum, $f\!\left(\tfrac{20}{3}\right) \approx 14{,}8$. $f''(x) = -6x + 26 = 0 \Rightarrow$ inflection at $x = \tfrac{13}{3} \approx 4{,}33$; slope of the tangent at the inflection $f'\!\left(\tfrac{13}{3}\right) = \tfrac{49}{3} \approx 16{,}33$.

(c) The graph is a cubic decreasing at its ends, with a minimum at $(2,-36)$, an inflection at $x=\tfrac{13}{3}$ and a maximum at $\left(\tfrac{20}{3},\,\approx 14{,}8\right)$.

(d) Inflection at $P(1,2)$, extremum at $x=0$. Since $f(0)=0$ automatically and $f'(0)=c$, the extremum at $0$ gives $c=0$. Inflection at $x=1$: $f''(1) = 6a + 2b = 0 \Rightarrow b = -3a$. Passing through $(1,2)$: $a + b + c = -2a = 2 \Rightarrow a = -1,\ b = 3,\ c = 0$. Moreover $f''(0) = 2b = 6 > 0$: at $x=0$ there is a minimum. Hence $f(x) = -x^3 + 3x^2$.

(e) Maximum at $P(-3,3)$, minimum at $x=0$. The extremum at $0$ gives $c = 0$, with $f''(0) = 2b > 0$ (minimum). Maximum at $x=-3$: $f'(-3) = 27a - 6b = 0 \Rightarrow b = \tfrac{9a}{2}$. Passing through $(-3,3)$: $-27a + 9b = \tfrac{27a}{2} = 3 \Rightarrow a = \tfrac{2}{9},\ b = 1,\ c = 0$.

(f) Inflection at $P(1,1)$ with horizontal tangent. Conditions $f'(1) = 0,\ f''(1) = 0,\ f(1) = 1$. From $f''(1) = 6a + 2b = 0 \Rightarrow b = -3a$; from $f'(1) = 3a + 2b + c = 0 \Rightarrow c = 3a$; from $f(1) = a + b + c = a = 1 \Rightarrow a = 1,\ b = -3,\ c = 3$. Hence $f(x) = x^3 - 3x^2 + 3x$. $\boxed{\text{(d) } a=-1,\,b=3,\,c=0;\quad \text{(e) } a=\tfrac29,\,b=1,\,c=0;\quad \text{(f) } a=1,\,b=-3,\,c=3}$