Q.Determine the intervals of concavity/convexity of the curve $y = x^3 - 3x + 1$ and hence find the point of inflection.

$f''(x)=6x$ changes sign at $x=0$, giving a point of inflection at $x=0$.

A point of inflection is where the second derivative is zero and changes sign; here that happens at $x=1$, giving $y=0$.

1. $y=(x-1)^3$. First derivative: $y'=3(x-1)^2$.
2. Second derivative: $y''=6(x-1)$.
3. Set $y''=0$: $6(x-1)=0\Rightarrow x=1$. …

The second derivative of $y=(x-1)^3$ vanishes and changes sign at $x=1$, giving the point of inflection $(1,0)$.

1. Let $y=(x-1)^3$.
2. Differentiating, $y'=3(x-1)^2$.
3. Differentiating again, $y''=6(x-1)$.
4. A point of inflection requires $y''=0$: $6(x-1)=0\Rightarrow x=1$. …

The curve $y=x^{1/3}$ has its point of inflection at $x=0$, precisely where $y''$ fails to exist.

1. Differentiate $y=x^{1/3}$: $y'=\dfrac{1}{3}x^{-2/3}$.
2. Differentiate again: $y''=\dfrac{1}{3}\cdot\left(-\dfrac{2}{3}\right)x^{-5/3}=-\dfrac{2}{9}x^{-5/3}$.
3. At $x=0$, $x^{-5/3}$ is undefined (division by zero), so $y''$ does not exist at $x=0$.
4. For $x>0$, $y''=-\dfrac{2}{9}x^{-5/3}<0$ (concave down); for $x<0$, $x^{-5/3}<0$ so $y''=-\dfrac{2}{9}x^{-5/3}>0$ (concave up). …

At a point of inflection (with the second derivative existing), the necessary condition is $f''(x_0)=0$.

1. A point of inflection is a point where the curve changes concavity — from concave-up to concave-down or vice versa.
2. Since the sign of $f''$ determines concavity, $f''$ must pass through zero (change sign) at the inflection point $x_0$.
3. Given that $f''(x_0)$ exists and changes sign there, the necessary condition is $f''(x_0)=0$. …