Parametric study of x³+2x²−ax

Text

Let $f(x) = x^3 + 2x^2 - ax$, with $a$ a real parameter.

• (a) compute $f'(x)$ keeping $a$ as a parameter;
• (b) for $a = 15$, study the sign of $f$ and find the coordinates of the relative maximum and minimum;
• (c) for which values of $a$ does the function have neither relative maxima nor minima?

Solution

(a) Derivative. $f'(x) = 3x^2 + 4x - a$.

(b) Case $a = 15$. The function becomes $f(x) = x^3 + 2x^2 - 15x = x(x+5)(x-3),$ with zeros $x = -5,\ 0,\ 3$; positive on $(-5,0)\cup(3,+\infty)$, negative elsewhere. The derivative is $f'(x) = 3x^2 + 4x - 15 = 0$, i.e. $x = \frac{-4 \pm \sqrt{16 + 180}}{6} = \frac{-4 \pm 14}{6} \ \Rightarrow\ x = -3 \ \text{or}\ x = \tfrac{5}{3}.$ At $x=-3$ the derivative changes from $+$ to $-$: maximum with $f(-3) = -27 + 18 + 45 = 36$. At $x=\tfrac{5}{3}$ the derivative changes from $-$ to $+$: minimum with $f\!\left(\tfrac{5}{3}\right) = -\tfrac{400}{27} \approx -14{,}81$.

(c) No relative extrema. We need $f'(x) = 3x^2 + 4x - a$ to have no two distinct real roots, i.e. discriminant $\le 0$: $\Delta = 16 + 12a \le 0 \ \Rightarrow\ a \le -\tfrac{4}{3}.$ $\boxed{\text{max } (-3,\,36);\quad \text{min } \left(\tfrac53,\,-\tfrac{400}{27}\right);\quad \text{no extrema for } a \le -\tfrac43}$