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Question
Show that the function f defined as follows, is continuous at x = 2, but not differentiable thereat:
Solution
Given:
First , we will show that f(x) is continuos at
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
Thus,
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)
\[ = \lim_{h \to 0} \frac{f(2 - h) - f(2)}{- h} \]
\[ = \lim_{h \to 0} \frac{2 h^2 - 7h + 6 - 6}{- h} \]
\[ = \lim_{h \to 0} - 2h + 7 \]
\[ = 7\]
(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.
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