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Find the value of k in the following so that the function f is continuous at the indicated point $f(x)=\left \{\begin{array}{1 1}3x-8, & if\;x\neq 5\\2k, & if\;x=5\end{array}\right.\;at\;x=5$

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Toolbox:
  • A function is said to be continuous at a point $'a'$ if the LHL = RHL.
  • (i.e) $\lim\limits_{\large x\to a^+}f(x)=\lim\limits_{\large x\to a^-}f(x)$
Step 1:
Given : $f(x)=\left \{\begin{array}{1 1}3x-8, & if\;x\neq 5\\2k, & if\;x=5\end{array}\right.\;at\;x=5$
The left hand limit :
$\lim\limits_{\large x\to 5}f(x)=\lim\limits_{\large x\to 5}f(5-h)$
$\qquad\qquad=\lim\limits_{\large x\to 5}3(5-h)-8$
$\qquad\qquad=\lim\limits_{\large x\to 5}15-3h-8$
Now applying the limits we get,
$\qquad\qquad=7$
Step 2:
It is given that the function is continuous at $x=5$
LHL=$2k$
$7=2k$
$\Rightarrow k=\large\frac{7}{2}$
Therefore the value of $k=\large\frac{7}{2}$
answered Jun 26, 2013 by sreemathi.v
 

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