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Find the values of \(k\) so that the function \(f\) is continuous at the indicated point in $ f(x) = \left\{ \begin{array} {1 1} kx+1 ,& \quad\text{ if $ x $ \(\leq 5\)}\\ 3x-5,& \quad \text{if $x$ > 5}\\ \end{array} \right.\qquad at \; x = 5 $

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  • If $f$ is a real function on a subset of the real numbers and $c$ be a point in the domain of $f$, then $f$ is continuous at $c$ if $\lim\limits_{\large x\to c} f(x) = f(c)$.
Step 1:
LHL=$\lim\limits_{\large x\to 5^-}(kx+1)=5k+1$
$f(5)=k.5+1$
$\quad\;\;\;=5k+1$
Step 2:
RHL=$\lim\limits_{\large x\to 5^+}(3x-5)$
$\quad\;\;\;=(3\times 5-5)$
$\quad\;\;\;=10$
Step 3:
$f$ is continuous if $LHL=RHL=f(5)$
$5k+1=10$
$5k=10-1$
$5k=9$
$k=\large\frac{9}{5}$
answered May 29, 2013 by sreemathi.v
 

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