f(x) = (x - 1)|x - 1|
\(\because\) x - 1 is a linear polynomial which is a continuous function.
Also |x - 1| is modulus function which is continuous function.
\(\therefore\) (x - 1)|x - 1| is also a continuous function
(\(\because\) If f and g are continuous function then f(x).g(x) is also continuous function)
Now, f(x) = (x - 1) |x - 1|
\(=\begin{cases}-(x-1)^2&;x\leq1\\(x-1)^2&;x\geq 1\end{cases}\)
\(\therefore\) f'(x) \(=\begin{cases}-2(x-1)&;x\leq1\\2(x-1)&;x\geq 1\end{cases}\)
f'(1 - h) = \(\lim\limits_{h\to 0}\)-2(1 - h - 1)
= \(\lim\limits_{h\to 0}\) 2h = 0
f'(1 + h) = \(\lim\limits_{h\to 0}\) 2(1 + h - 1) = \(\lim\limits_{h\to 0}\) 2h = 0
\(\because\) f'(1 - h) = f'(1 + h)
\(\therefore\) f(x) is differentiable x = 1
\(\therefore\) f(x) is a continuous and differentiable function at x = 1.
