snowlovelydayM

2021-09-03

$f\left(x\right)=9,\phantom{\rule{1em}{0ex}}\text{if}\phantom{\rule{1em}{0ex}}x\le -4,\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}f\left(x\right)=ax+b,\phantom{\rule{1em}{0ex}}\text{if}\phantom{\rule{1em}{0ex}}-4
Find the constants, if the function is continuous on the entire line.

unett

The function is continuous, if
$\underset{x\to {a}^{+}}{lim}f\left(x\right)=\underset{x\to {a}^{-}}{lim}f\left(x\right)=f\left(a\right)$
All of the given functions are continous, since $\underset{1}{f}\left(x\right),\underset{3}{f}\left(x\right)$ are constant function and $\underset{2}{f}\left(x\right)$ is linear function. We need to check only at point where the function might break (-4 and 5):
At $x=-4$
$\underset{x\to -{4}^{-}}{lim}f\left(x\right)=\underset{x\to -{4}^{-}}{lim}9$
=9
$\underset{x\to -{4}^{+}}{lim}f\left(x\right)=\underset{x\to -{4}^{-}}{lim}ax+b$
$=a\left(-4\right)+b$
$=-4a+b$
$f\left(-4\right)=9$
$\underset{x\to -{4}^{-}}{lim}f\left(x\right)=\underset{x\to -{4}^{+}}{lim}f\left(x\right)=f\left(-4\right)$
$9=-4a+b=9$
$-4a+b=9$ (I)
At $x=5$
$\underset{x\to {5}^{+}}{lim}f\left(x\right)=\underset{x\to {5}^{-}}{lim}ax+b$
$=5a+b$
$\underset{x\to {5}^{+}}{lim}f\left(x\right)=\underset{x\to {5}^{+}}{lim}-9$
$=-9$
$f\left(5\right)=-9$
If the condition of continuity is satisfied,
$\underset{x\to {5}^{-}}{lim}f\left(x\right)=\underset{x\to {5}^{+}}{lim}f\left(x\right)=f\left(5\right)$
$5a+b=-9$ (II)
Subtract (II) from (I) to get:
$-9a+0=18$
$\to a=-2$
Substitute $a=-2$ in (I)
$-4\left(-2\right)+b=9$
$8+b=9$
$b=1$
$a=-2$
$b=1$