Cubic, inflection, concavity and monotonicity

Text

Given the function $y = x^3 - 3x^2 - 2x + 1$:

• (a) determine where the inflection point is, for which $x$ the graph is concave down and for which it is concave up;
• (b) determine for which $x$ the function is increasing and for which it is decreasing.

Solution

(a) Concavity and inflection. $y' = 3x^2 - 6x - 2$ and $y'' = 6x - 6$. $y''=0 \iff x=1$. For $x<1$ we have $y''<0$ (concave down), for $x>1$ we have $y''>0$ (concave up): at $x=1$ the concavity changes, so it is an inflection point.

(b) Monotonicity. $y' = 3x^2 - 6x - 2 = 0$ at $x = \frac{6 \pm \sqrt{36 + 24}}{6} = \frac{6 \pm \sqrt{60}}{6} = 1 \pm \frac{\sqrt{15}}{3} \approx -0.29 \ \text{and}\ 2.29.$ Since $y'$ is an upward parabola, the function is increasing for $x < 1-\tfrac{\sqrt{15}}{3}$ and $x > 1+\tfrac{\sqrt{15}}{3}$, decreasing between the two values. $\boxed{\text{inflection at } x=1;\quad \text{concave up for } x>1,\ \text{concave down for } x<1}$