Show that function ƒ(x) = {(1-x), when x < 1; (x^2-1), when x ≥1. is continuous but not differentiable at x=1

Question

Show that function  \(f(x) = \begin{cases} (1-x), & \quad \text{when} \,x<1; \text{}\\ (x^2-1), & \quad \text{when} \,x≥1 \text{} \end{cases}\)   is continuous but not differentiable at x=1

Answer 1

Given function \(f(x) = \begin{cases} (1-x), & \quad \text{when} \,x<1; \text{}\\ (x^2-1), & \quad \text{when} \,x≥1 \text{} \end{cases}\) 

Left hand limit at x = 1: 

\(\lim\limits_{x \to 1^-}\) f(x) = \(\lim\limits_{x \to 1}\)  (1-x) = 1 - 1 = 0

Right hand limit at x = 1: 

 \(\lim\limits_{x \to 1^+}\) f(x) = \(\lim\limits_{x \to 1}\)  (x²- 1) = 1² - 1 = 0

Also, f(1) = 1² – 1 = 0 

As, 

Therefore, 

f(x) is continuous at x = 1 

Now, let’s see the differentiability of f(x):

 LHD at x = 2:

As, LHD ≠ RHD 

Therefore, f(x) is not differentiable at x = 2