sagnuhh

2021-09-10

Consider that
$f\left(x\right)=k+6x,\phantom{\rule{1em}{0ex}}\text{if}\phantom{\rule{1em}{0ex}}x\le -2\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}f\left(x\right)=kx+9,\phantom{\rule{1em}{0ex}}\text{if}\phantom{\rule{1em}{0ex}}x\ge -2$
and find the value of k.

krolaniaN

$\underset{x\to -{2}^{-}}{lim}f\left(x\right)=\underset{x\to -{2}^{-}}{lim}k+6x$
$=k+6\left(-2\right)$
$=k-12$
$\underset{x\to -{2}^{+}}{lim}f\left(x\right)=\underset{x\to -{2}^{+}}{lim}kx+9$
$=k\left(-2\right)+9$
$=-2k+9$
For the function to be continuous:
$\underset{x\to -{2}^{-}}{lim}f\left(x\right)=\underset{x\to -{2}^{+}}{lim}f\left(x\right)$
Thus,
$k-12=-2k+9$
$3k=21$
$k=7$

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