Show that the function f(x) = |x - 3 |,x |R,is continuous but not differentiable at x = 3.

Here, f(x) =|x - 3| f(x) - (x - 3 ) ,x 3 Now, * lim _ x 3^ + f(x) = * lim _ h 0 f(3 + h) [Let x = 3 + h and x 3^ + 0] * lim _ h 0 (3 + h - 3 ) = * lim _ h 0 h = 0 * lim _ x 3^ + f(x) = 0 - - - - - (i) * lim _ x 3^ - f(x) = * lim _ h 0 f(3 - h) [Let x = 3 - h and xx 3^ - 0] = * lim _ h 0 - (3 - h - 3) = * lim _ h 0 =h = 0 = * lim _ x3^ + f(x) - - - -- (ii) Also, f(3) = 0 - - - --- (iii) From equation (i), (ii) and (iii) = * lim _ x3^ + f(x) = * lim _ x3^ - f(x) =f(3) Hence, f(x) is continuous at x = 3 At x = 3 RHD = * lim _ h 0 f(3 + h) - f(3) h = * lim _ h 0 (3 + h - 3)- 0 h = * lim _ h 0 h h [ |h|= h,|0|= 0] = * lim _ h 0 1 RHD = 1 - - - - - - - (iv) LHD = * lim _ h 0 f(3 - h)- f(3) -h = * lim _ h 0 -(3-h-3)-0 -h = * lim _ h 0 h -h [ |h|= h] = * lim _ h 0 (-1) LHD = – 1 - - - - - - -(v) Equation (iv) and (v) RHD LHD at x = 3. Hence f(x) is not differentiable at x = 3 Therefore, f(x) =|x - 3|, x R is continuous but not differentiable at x = 3.

Show that the function f(x) = |x - 3 |,x |R,is continuous but not…

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Show that the function $\text{f}(\text{x}) = |\text{x} - 3 |,\text{x}\in|\text{R},$is continuous but not differentiable at x = 3.

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Here, f(x) =|x - 3|
$\text{f}(\text{x}) - (\text{x} - 3 ) ,\text{x} < 3 $
$0 ,\text{x} = 3 $
$(\text{x} - 3 ),\text{x} > 3 $
Now, $\DeclareMathOperator*{\median}{\text{lim}} \median_{\text{x}\rightarrow3^{+}}\text{f}(\text{x}) = \DeclareMathOperator*{\median}{\text{lim}} \median_{\text{h}\rightarrow0}\text{f}(3 + \text{h})$
[Let x = 3 + h and $\text{x}\rightarrow3^{+}\Rightarrow\text{h}\rightarrow0]$
$\DeclareMathOperator*{\median}{\text{lim}} \median_{\text{h}\rightarrow0}(3 + \text{h} - 3 ) = \DeclareMathOperator*{\median}{\text{lim}} \median_{\text{h}\rightarrow0}\text{h} = 0 $
$\DeclareMathOperator*{\median}{\text{lim}} \median_{\text{x}\rightarrow3^{+}}\text{f}(\text{x}) = 0 $ - - - - - (i)
$\DeclareMathOperator*{\median}{\text{lim}} \median_{\text{x}\rightarrow3^{-}}\text{f}(\text{x}) = \DeclareMathOperator*{\median}{\text{lim}} \median_{\text{h}\rightarrow0}\text{f}(3 - \text{h})$
[Let x = 3 - h and x$\text{x}\rightarrow3^{-}\Rightarrow\text{h}\rightarrow0]$
$ = \DeclareMathOperator*{\median}{\text{lim}} \median_{\text{h}\rightarrow0} - (3 - \text{h} - 3) = \DeclareMathOperator*{\median}{\text{lim}} \median_{\text{h}\rightarrow0}=\text{h} = 0$
$ = \DeclareMathOperator*{\median}{\text{lim}} \median_{\text{x}\rightarrow{3}^{+}} \text{f}(\text{x})$ - - - -- (ii)
Also, f(3) = 0 - - - --- (iii)
From equation (i), (ii) and (iii)
$ = \DeclareMathOperator*{\median}{\text{lim}} \median_{\text{x}\rightarrow{3}^{+}} \text{f}(\text{x}) = \DeclareMathOperator*{\median}{\text{lim}} \median_{\text{x}\rightarrow{3}^{-}} \text{f}(\text{x}) =\text{f}(3)$
Hence, f(x) is continuous at x = 3
At x = 3
RHD $ = \DeclareMathOperator*{\median}{\text{lim}} \median_{\text{h}\rightarrow0} \frac{\text{f}(3 + \text{h}) - \text{f}(3)}{\text{h}} = \DeclareMathOperator*{\median}{\text{lim}} \median_{\text{h}\rightarrow0}\frac{(3 + \text{h} - 3)- 0}{\text{h}}$
$ = \DeclareMathOperator*{\median}{\text{lim}} \median_{\text{h}\rightarrow0} \frac{\text{h}}{\text{h}}$ [$\because$|h|= h,|0|= 0]
$ = \DeclareMathOperator*{\median}{\text{lim}} \median_{\text{h}\rightarrow0}1$
RHD = 1 - - - - - - - (iv)
LHD $ = \DeclareMathOperator*{\median}{\text{lim}} \median_{\text{h}\rightarrow0}\frac{\text{f}(3 - \text{h})- \text{f}(3)}{-\text{h}} = \DeclareMathOperator*{\median}{\text{lim}} \median_{\text{h}\rightarrow0} \frac{-(3-\text{h}-3)-0}{-\text{h}}$
$ = \DeclareMathOperator*{\median}{\text{lim}} \median_{\text{h}\rightarrow0}\frac{\text{h}}{-\text{h}}$ [$\because$|h|= h]
$ = \DeclareMathOperator*{\median}{\text{lim}} \median_{\text{h}\rightarrow0}(-1)$
LHD = – 1 - - - - - - -(v)
Equation (iv) and (v) $\Rightarrow$RHD $\neq$LHD at x = 3.
Hence f(x) is not differentiable at x = 3
Therefore, f(x) =|x - 3|, x$\in$R is continuous but not differentiable at x = 3.

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