If f(x)=(x^2+6x+8)/(x+2) if x<-2 and kx^2if x>= -2, what value of k would make the function continuous at x=-2?

Jul 18, 2016

$k = \frac{1}{2}$.

Explanation:

For $x < - 2 , f \left(x\right) = \frac{{x}^{2} + 6 x + 8}{x + 2} = \frac{\left(x + 4\right) \left(x + 2\right)}{x + 2}$, and, since,

$x \ne - 2$, we can cancel $\left(x + 2\right)$ to get,

$f \left(x\right) = \frac{\left(x + 4\right) \cancel{x + 2}}{\cancel{x + 2}} = x + 4 , \mathmr{if} , x < - 2$

Now, as $x \rightarrow - 2 - , x < - 2 , s o , f \left(x\right) = x + 4$

$\therefore {\lim}_{x \rightarrow - 2 -} f \left(x\right) = {\lim}_{x \rightarrow - 2 -} \left(x + 4\right) = - 2 + 4 = 2. \ldots \ldots \ldots \left(1\right)$

Further, as $x \rightarrow - 2 + , x \succ 2 , s o , f \left(x\right) = k {x}^{2}$

:. lim_(xrarr-2+) f(x)=lim_(xrarr-2+)kx^2=4k...................(2)

Also, $f \left(- 2\right) = 4 k \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \left(3\right)$

Because $f$ is reqd. to be continuous at $x = - 2$, by defn. of continuity, we must have,

${\lim}_{x \rightarrow - 2 -} f \left(x\right) = f \left(- 2\right) = {\lim}_{x \rightarrow - 2 +} f \left(x\right)$.

$\left(1\right) , \left(2\right) \mathmr{and} \left(3\right) \Rightarrow 2 = 4 k \Rightarrow k = \frac{1}{2}$.