Given the piecewise function  1-x,x<-1, (x^2)-x,-1<=x<=6, x-7,x>6, is it continuous at x=-1 and -6?

Dec 22, 2015

It is continuous at $x = - 1$ and $x = - 6$, but not at $x = 6$.

Explanation:

*As a disclaimer, I assumed the $- 6$ at the very end of the question was supposed to be a $6$ and solved accordingly. *

To solve, you simply plug in the bordering $x$-value and see if the two are the same.

For $x = - 1$:
$1 - \left(- 1\right)$
$1 + 1$
$2$

${\left(- 1\right)}^{2} - \left(- 1\right)$
$1 + 1$
$2$

So, $f \left(x\right)$ is continuous at $x = - 1$

For $x = - 6$, the function is continuous because it is not a border of the piecewise.

For $x = 6$:
${\left(6\right)}^{2} - \left(6\right)$
$36 - 6$
$30$

$\left(6\right) - 7$
$- 1$
So, $f \left(x\right)$ is not continuous at $x = 6$