# Find the values of a and b such that

Question:

Find the values of and such that the function defined by

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

is a continuous function at = 4.

Solution:

Finding the left hand and right hand limit for the given function, we have

$\lim _{x \rightarrow 4} f(x)=\frac{x-4}{|x-4|}+a$

$=\lim _{h \rightarrow 0} \frac{4-h-4}{|4-h-4|}+a=\lim _{h \rightarrow 0} \frac{-h}{h}+a=-1+a$

$\lim _{x \rightarrow 4} f(x)=a+b$

$\lim _{x \rightarrow 4^{-}} f(x)=\frac{x-4}{|x-4|}+b$

$=\lim _{h \rightarrow 0} \frac{4+h-4}{|4+h-4|}+b=\lim _{h \rightarrow 0} \frac{h}{h}+b=1+b$

As the function is continuous at $x=4$.

$\therefore \quad \lim _{x \rightarrow 4^{-}} f(x)=\lim _{x \rightarrow 4} f(x)=\lim _{x \rightarrow 4^{-}} f(x)$

So, -1 + a = a + b = 1 + b

-1 + a = a + b and 1 + b = a + b

We get, b = -1 and 1 + -1 = a + -1 ⇒ a = 1

Therefore, the value of a = 1 and b = -1