Question

If \[f\left( x \right) = \begin{cases}\frac{x - 4}{\left| x - 4 \right|} + a, \text{ if }  & x < 4 \\ a + b , \text{ if } & x = 4 \\ \frac{x - 4}{\left| x - 4 \right|} + b, \text{ if } & x > 4\end{cases}\]  is continuous at x = 4, find a, b.

 

Answer 1

Given: 

 \[f\left( x \right) = \begin{cases}\frac{x - 4}{\left| x - 4 \right|} + a, \text{ if }  & x < 4 \\ a + b , \text{ if } & x = 4 \\ \frac{x - 4}{\left| x - 4 \right|} + b, \text{ if } & x > 4\end{cases}\] 

We observe
(LHL at x = 4) = 

\[\lim_{x \to 4^-} f\left( x \right) = \lim_{h \to 0} f\left( 4 - h \right)\]

\[= \lim_{h \to 0} \left( \frac{4 - h - 4}{\left| 4 - h - 4 \right|} + a \right) = \lim_{h \to 0} \left( \frac{- h}{\left| - h \right|} + a \right) = a - 1\]

(RHL at x = 4) = 

\[\lim_{x \to 4^+} f\left( x \right) = \lim_{h \to 0} f\left( 4 + h \right)\]

\[= \lim_{h \to 0} \left( \frac{4 + h - 4}{\left| 4 + h - 4 \right|} + b \right) = \lim_{h \to 0} \left( \frac{h}{\left| h \right|} + b \right) = b + 1\]

And

\[f\left( 4 \right) = a + b\]

If f(x) is continuous at x = 4, then 

\[\lim_{x \to 4^-} f\left( x \right) = \lim_{x \to 4^+} f\left( x \right) = f\left( 4 \right)\]

\[\Rightarrow a - 1 = b + 1 = a + b\]

\[\Rightarrow a - 1 = a + b, b + 1 = a + b\]

\[\Rightarrow b = - 1, a = 1\]