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Answer the following.
Prove that the function f(x) = |x - 1|, x ∈ mathbbR, is continuous at x = 1 but not differentiable.
Prove that the function $f(x) = |x - 1|, x \in \mathbb{R}$, is continuous at $x = 1$ but not differentiable.
Given: $f(x) = |x - 1|, x \in \mathbb{R}$
To show: $f(x)$ is continuous at $x = 1$ but not differentiable
Answer
Answer
$f(x) = |x-1| = \begin{cases} x-1, & x \ge 1 \\ 1-x, & x < 1 \end{cases}$
Continuity at $x=1$: LHL $= \lim_{x\to1^-}(1-x) = 0$, RHL $= \lim_{x\to1^+}(x-1) = 0$, $f(1) = 0$.
LHL $=$ RHL $= f(1)$, so $f$ is continuous at $x=1$.
LHD $= \lim_{h\to0^-}\frac{f(1+h)-f(1)}{h} = \frac{-h}{h} = -1$; RHD $= \lim_{h\to0^+}\frac{h}{h} = 1$.
LHD $\ne$ RHD.
Hence $f$ is continuous but not differentiable at $x=1$.
- Write $f(x) = 1-x$ for $x<1$ and $f(x) = x-1$ for $x \ge 1$, by definition of modulus.
- LHL $= \lim_{x\to1^-}(1-x) = 0$ and RHL $= \lim_{x\to1^+}(x-1) = 0$.
- $f(1) = 0 = $ LHL $=$ RHL, so $f$ is continuous at $x=1$.
- LHD $= \lim_{h\to0^-}\frac{|h|-0}{h} = \lim \frac{-h}{h} = -1$.
- RHD $= \lim_{h\to0^+}\frac{|h|-0}{h} = \lim\frac{h}{h} = 1$.
- LHD $\ne$ RHD, so $f$ is not differentiable at $x=1$.