Study of 4+2x+√(x²−x−6)

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Given the function $f(x) = 4 + 2x + \sqrt{x^2 - x - 6}$:

• find the domain;
• study the sign;
• compute the limits at the boundaries of the domain;
• find any oblique asymptotes;
• determine the relative maximum/minimum points (showing that one lies at $x \approx -2{,}39$).

Solution

Domain. We need $x^2 - x - 6 \ge 0$, i.e. $(x-3)(x+2) \ge 0 \Rightarrow x \le -2 \ \lor\ x \ge 3$.

Boundary values. $f(-2) = 4 - 4 + 0 = 0$ and $f(3) = 4 + 6 + 0 = 10$.

Sign. On the domain $f(x) \ge 0$ throughout: for $x \ge 3$ all terms are positive; for $x \le -2$ one checks that $4 + 2x + \sqrt{x^2 - x - 6} \ge 0$ (equality only at $x=-2$).

Oblique asymptotes. As $x \to +\infty$, $\sqrt{x^2 - x - 6} \sim x - \tfrac12$, so $f \sim 3x + \tfrac72$: asymptote $y = 3x + \tfrac72$. As $x \to -\infty$, $\sqrt{x^2 - x - 6} = |x|\sqrt{1 - \tfrac1x - \tfrac{6}{x^2}} \sim -x + \tfrac12$, so $f \sim x + \tfrac92$: asymptote $y = x + \tfrac92$.

Extrema. $f'(x) = 2 + \dfrac{2x - 1}{2\sqrt{x^2 - x - 6}}$. Setting $f'(x) = 0$ on the branch $x \le -2$: $4\sqrt{x^2 - x - 6} = -(2x - 1) \ \Rightarrow\ 16(x^2 - x - 6) = (2x-1)^2 \ \Rightarrow\ 12x^2 - 12x - 97 = 0,$ giving $x = \dfrac{12 \pm \sqrt{144 + 4656}}{24}$, i.e. $x \approx -2{,}387$ (acceptable, $\le -2$) and $x \approx 3{,}39$ (rejected, it does not annul the derivative on the branch $x \ge 3$). The relative extremum is therefore at $\approx (-2{,}39,\ f(-2{,}39))$. $\boxed{\text{Domain } x \le -2 \lor x \ge 3;\quad y = x + \tfrac92 \ (x\to-\infty),\ y = 3x + \tfrac72 \ (x\to+\infty);\quad x_e \approx -2{,}387}$