JEE Advanced · 2006 · 09 Apr 2006 · Continuity and Differentiability

To find $f(x)=\min \{1,x^2,x^3\}$, we need to determine which function is smallest in different regions.

For $x\le 1$: We have $x^3\le x^2\le 1$ (for $x\in [0,1]$) and for $x<0$, $x^3$ is negative and smallest. So $f(x)=x^3$ for $x\le 1$.

For $x>1$: We have $1<x^2<x^3$, so $f(x)=1$ for $x>1$.

Therefore:


At $x=1$: ${\lim }_{x\to 1^{-}}f(x)=1^3=1$ and ${\lim }_{x\to 1^{+}}f(x)=1$, so $f$ is continuous at $x=1$ and hence continuous $\forall x\in \mathbb{R}$.

For differentiability at $x=1$:

• Left derivative: ${\lim }_{h\to 0^{-}}\frac{f(1+h)-f(1)}{h}={\lim }_{h\to 0^{-}}\frac{(1+h{)}^3-1}{h}=3$
• Right derivative: ${\lim }_{h\to 0^{+}}\frac{f(1+h)-f(1)}{h}={\lim }_{h\to 0^{+}}\frac{1-1}{h}=0$

Since left and right derivatives differ, $f$ is not differentiable at $x=1$ (only one point, not two).

Therefore, $f(x)$ is continuous everywhere but not differentiable at $x=1$.