Question

Let f(x) = {1-x, if x<=-1
{SQRT (x+b), if x>-1
Determine the value of b that makes f(x) continuous at x=-1.

Answer 1

\(f\) is continuous at -1 if
\[\lim_{x\to-1^-}f(x)=\lim_{x\to-1^+}f(x)\]
and these limits are equal to \(f(-1)\).

Answer 2

\[\lim_{x\to-1^-}f(x)=\lim_{x\to-1^-}(1-x)=1-(-1)=2\]
\[\lim_{x\to-1^+}f(x)=\lim_{x\to-1^+}\sqrt{x+b}=\sqrt{b-1}\]
By the definition of \(f\), you have \(f(-1)=2\), so the function is continuous at -1 if
\[\sqrt{b-1}=2\]

Answer 3

that's what I initially thought but pluged itin to the online course and it didn't work

Answer 4

You got \(b=5\), right?

Answer 5

now I now thanks for the help

Answer 6

yw