Find the values of k so that the function f is continuous at the … - Vidyadip

Question

Find the values of k so that the function f is continuous at the indicated point:
$\text{f(x)}=\begin{cases}\text{k}\text{x}+1,&\text{if}\ \text{x}\leq{5}\\3\text{x} - 5,& \text{if}\ \text{x}> 5\end{cases}$
$\text{at} \text{x} = 5$

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Answer

Here $\text{f(x)}=\begin{cases}\text{k}\text{x}+1,&\text{if}\ \text{x}\leq{5}\\3\text{x} - 5,& \text{if}\ \text{x}> 5\end{cases}$
$^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{5}^{-}}\text{f(x)}=^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{5}^{-}}(\text{k}\text{x} + 1) = 5\text{k} + 1$
$^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{5}^{+}}\text{f(x)}=^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{5}^{+}}({3}\text{x} - 5) = 15 - 5 = 10$
Since f is continuous at x = 1
$\therefore\ ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{5}^{-}}\text{f(x)}=^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{5}^{+}}\text{f(x)}$
$\therefore 5 \text{k} + 1 = 10 \Rightarrow 5\text{k} = 9 \Rightarrow\text{k} = \frac{9}{5}$

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