The function is $f(x)=x²+cx+6$ if $x<1$;$f(x)=x²-x-c$ if $x\geq1$. I approached the problem like this. For $f(x)$ is continuous if $\lim_{x\to1} f(x)=f(1)$
$\lim_{x\to1+} x²-x-c=-c=f(1)$. Also $\lim_{x\to1-}x²+cx+6=c+7$. Then $-c=c+7$ and therefore $c=-7/2$. Graphing I have realized that I'm wrong what am i missing or doing incorrectly?. Thanks a lot in advance!
You have $f(1)=-c$ and also $$ \lim_{x\to1^+}f(x)=-c $$ Moreover $$ \lim_{x\to1^-}f(x)=c+7 $$ So continuity requires $$ c+7=-c $$ hence $c=-7/2$. So this is right.