Consider the following in respect of the function f(x) = |x - 3| :

1. f(x) is continuous at x = 3

2. f(x) is differentiable at x = 0.

Which of the above statements is / are correct?

Concept:

f(x) is continuous at x = a, if \(\rm\lim _{x \rightarrow a^{-}} f(x)=\lim _{x \rightarrow a^{+}} f(x)=lim _{x \rightarrow a} f(x)\)

f(x) is differentiable at x = a, if LHD = RHD

 \(\begin{array}{l} \rm L H D=\lim _{h \rightarrow 0^{-}} \frac{f(a-h)-f(a)}{-h} \\ \rm R H D=\lim _{h \rightarrow 0^{+}} \frac{f(a+h)-f(a)}{h} \end{array}\)

f(x) = |x|, f(x) = x, if x > 0 and f(x) = -x, if x < 0

Calculation:

LHL = \(\rm\lim_{x\rightarrow3^-}f(x)=\lim_{x\rightarrow3^-}-(x-3)=0\)

RHL = \(\rm\lim_{x\rightarrow3^+}f(x)=\lim_{x\rightarrow3^+}(x-3)=0\)

f(x = 3) = 0

∴ f(x) is continuous at x = 3

\(\begin{array}{l} \rm L H D=\lim _{\rm h \rightarrow 0^{-}} \frac{f(0-0)-f(0)}{-h} \\ =\lim _{\rm h \rightarrow 0^{-}} \frac{f(0)-f(0)}{-h}\\=0\\ \rm R H D=\lim _{h \rightarrow 0^{+}} \frac{f(a+h)-f(a)}{h}\\ =0 \end{array}\)

LHD = RHD, so f(x) is differentiable at x = 0.

Hence, option (3) is correct.