Show that ƒ(x) is continuous but not differentiable at x=1 Let f(x) = {(2-x), when x ≥ 1; x, when 0 ≤x ≤1.

Question

Show that ƒ(x) is continuous but not differentiable at x=1

Let   \(f(x) = \begin{cases} (2-x), & \quad \text{when}\, x≥1; \text{}\\ x, & \quad \text{when} \,0 ≤x≤1 \text{} \end{cases}\) 

Answer 1

Left hand limit at x = 1

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

f(x) = x is polynomial function and a polynomial function is continuous everywhere 

Right hand limit at x = 1

 \(\lim\limits_{x \to 1^+}\) f(x) = \(\lim\limits_{x \to 1}\) (2-x) = (2-1) = 1

f(x) = 2 - x is polynomial function and a polynomial function is continuous everywhere 

Also, f(1) =1 

As we can see that, 

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

Therefore, 

f(x) is continuous at x =1

Now, 

LHD at x = 1

As, LHD ≠ RHD 

Therefore, 

f(x) is not differentiable at x = 1