Q.Find intervals of concavity and points of inflexion for the following functions:

$f''(x)=6x$ 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 $y''=0$: $6(x-1)=0\Rightarrow x=1$.
4. Check sign change: for $x<1$, $y''<0$ (concave down); for $x>1$, $y''>0$ (concave up) — concavity genuinely changes at $x=1$, confirming an inflection point.
5. At $x=1$: $y=(1-1)^3=0$.
6. So the point of inflection is $(1,0)$.

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 $y''=0$: $6(x-1)=0\Rightarrow x=1$.
5. Checking $y'''=6\ne0$ confirms $y''$ genuinely changes sign at $x=1$ (from negative for $x<1$ to positive for $x>1$), so this is a true inflection point.
6. At $x=1$: $y=(1-1)^3=0$.
7. So the point of inflection is $(1,0)$.

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).
5. Since concavity changes sign on either side of $x=0$, the curve has a genuine point of inflection at $x=0$, even though $y''(0)$ itself does not exist.
6. This rules out the option requiring $y''=0$ at the inflection point, since here $y''$ is simply undefined there.

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 $f''(x_0)=0$.
4. $f(x_0)=0$ describes a root of the function and $f'(x_0)=0$ describes a stationary point — neither is the defining condition for inflection; $f''(x_0)\neq0$ is the opposite of what inflection requires.