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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!

Andres Romero
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  • Your calculation is correct, you likely graphed it incorrectly. I just graphed it, and it looks fine. – 高田航 Apr 18 '17 at 22:19
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    I just realized my mistake ignoring a negative sign. Good think is that I learned a little bit more about writing mathematical equations and symbols in LaTeX. Thanks – Andres Romero Apr 18 '17 at 22:23

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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.

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egreg
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