Show that the function f defined as follows, is continuous at x =… - Vidyadip

Question

Show that the function f defined as follows, is continuous at x = 2, but not differentiable thereat:
f(x) = $\begin{matrix} \text{3x - 2}, && 0<x\leq1 \\ \text{ 2x}^{2}-\text{x}, && 1<x\leq2 \\ \text{5x - 4}, && x>2 \end{matrix}$.

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Answer

$\DeclareMathOperator*{\median}{\text{lim}} \median_{\text{h}\rightarrow0}\text{ f(2 - h)}=\DeclareMathOperator*{\median}{\text{lim}} \median_{\text{h}\rightarrow0}[2(2 - \text{h})^{2}-(2 - \text{h})]=6\ .......\text{(i)}$
$\DeclareMathOperator*{\median}{\text{lim}} \median_{\text{h}\rightarrow0}\text{ f(2 + h)}=\DeclareMathOperator*{\median}{\text{lim}} \median_{\text{h}\rightarrow0}[5(2 + \text{h})-4 ]=6\ ............\text{(ii)}$
f(2) = 8 – 2 = 6 ...........(iii)
From (i), (ii), and (iii), f(x) is continuous at x = 2
$\text{RHD}=\DeclareMathOperator*{\median}{\text{lim}} \median_{\text{h}\rightarrow0}\Bigg[\frac{\left\{5{(2+h)-4}-(6)\right\}}{h}\Bigg]\neq\text{LHD}=\DeclareMathOperator*{\median}{\text{lim}} \median_{\text{h}\rightarrow0}\Bigg[\frac{\left\{(2h - 3)(h - 2) - 6\right\}}{-h}\Bigg]\text{as 5}\neq7$
$\therefore$ f(x) is not differentiable there at.

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