Question

Need help with this continuity problem.

Answer 1

In order to be continuous everywhere, we only need to satisfy continuity at $x=10$. For that to happen, we need to have

$\displaystyle\lim_{x\to10^-}f(x)=\lim_{x\to10^+}f(x)=f(10)$

By the function's definition, $f(10)=2(10)-2=18$. Compute the limits:

$\displaystyle\lim_{x\to10^-}f(x)=\lim_{x\to10}2x-2=18$

$\displaystyle\lim_{x\to10^+}f(x)=\lim_{x\to10}-8x+b=-80+b$

So we must have

$18=b-80\implies b=98$