Question

Show that the function f defined as follows, is continuous at x = 2, but not differentiable thereat: 

\[f\left( x \right) = \begin{cases}3x - 2, & 0 < x \leq 1 \\ 2 x^2 - x, & 1 < x \leq 2 \\ 5x - 4, & x > 2\end{cases}\]

Answer 1

Given:  

\[f\left( x \right) = \begin{cases}3x - 2, & 0 < x \leq 1 \\ 2 x^2 - x, & 1 < x \leq 2 \\ 5x - 4, & x > 2\end{cases}\]

First , we will show that f(x) is continuos at 

\[x = 2\]

We have,
(LHL at x=2)

\[{= \lim}_{x \to 2^-} f(x)\]
\[ = \lim_{h \to 0} f(2 - h) \]
\[ = \lim_{h \to 0} 2(2 - h )^2 - (2 - h)\]
\[ = \lim_{h \to 0} (8 + 2 h^2 - 8h - 2 + h)\]
\[ = 6\]

(RHL at x = 2) 

\[= \lim_{x \to 2^+} f(x) \]
\[ = \lim_{h \to 0} f(2 + h) \]
\[ = \lim_{h \to 0} 5(2 + h) - 4 \]
\[ = \lim_{h \to 0} (10 + 5h - 4) \]
\[ = 6\

and 

\[f(2) = 2 \times 4 - 2 = 6 .\]

Thus,  

\[\lim_{x \to 2^-} f(x)\]

\[\lim_{x \to 2^+} f(x)\]

\[f(2)\]

Hence the function is continuous at x=2.

Now, we will check whether the given function is differentiable at x = 2.

We have,
(LHD at x = 2)

\[\lim_{x \to 2^-} \frac{f(x) - f(2)}{x - 2} \]
\[ = \lim_{h \to 0} \frac{f(2 - h) - f(2)}{- h} \]
\[ = \lim_{h \to 0} \frac{2 h^2 - 7h + 6 - 6}{- h} \]
\[ = \lim_{h \to 0} - 2h + 7 \]
\[ = 7\]

(RHD at x = 2)

\[\lim_{x \to 2^+} \frac{f(x) - f(2)}{x - 2} \]
\[ = \lim_{h \to 0} \frac{f(2 + h) - f(2)}{h} \]
\[ = \lim_{h \to 0} \frac{10 + 5h - 4 - 6}{h}\]
\[ = 5\]

Thus, LHD at x=2 ≠ RHD at x = 2.
Hence, function is not differentiable at x = 2.