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`