Find the relationship between a and b so that the function f defined by

$f(x) = \left\{ {\begin{array}{llllllllllllllllllll}{ax + 1,}&{if}&{x \le 3}\\{bx + 3,}&{if}&{x > 3}\end{array}} \right.$ is continuous at x$=$ 3.

At x$=$ 3:

$\mathop {\lim }\limits_{x \to {3^ - }} f(x) = \mathop {\lim }\limits_{x \to {3^ - }} (ax + 1) = \mathop {\lim }\limits_{\scriptstyle x \to 3 - h\atop\scriptstyle h \to 0} (a(3 - h) + 1)$

$= \mathop {\lim }\limits_{x \to 3 - h} (3a - ah + 1) = 3a + 1$

$\mathop {\lim }\limits_{x \to {3^ + }} f(x) = \mathop {\lim }\limits_{x \to {3^ - }} (bx + 3) = \mathop {\lim }\limits_{\scriptstyle x \to 3 + h\atop\scriptstyle h \to 0} (b(3 + h) + 3)$

$= \mathop {\lim }\limits_{\scriptstyle x \to 3 + h\atop\scriptstyle h \to 0} (3b + bh + 3) = 3b + 3$
Also, $f(3) = 3a + 1$

Thus, $\mathop {\lim }\limits_{x \to {3^ - }} f(x) = \mathop {\lim }\limits_{x \to {3^ + }} f(x) = f(3)$

f(x) is given as continuous at $x = 3]$

$\Rightarrow$ $3b + 3 = 3a + 1 \Rightarrow 2 = 3(a - b) \Rightarrow a - b = \cfrac{2}{3}$

Therefore, this is the required relation between a and b.