Find the values of $$a$$ and $$b$$ such that the function defined by $f(x) = \left\{ \begin{array} {1 1} 5 ,& \quad\text{ if$ x $$$\leq 2$$}\\ ax+b,& \quad \text{if$2$< x<10}\\ 21,& \quad \text{if$ x $$$\geq 10$$}\\ \end{array} \right.$ is a continuous function.

Toolbox:
• If $f$ is a real function on a subset of the real numbers and $c$ be a point in the domain of $f$, then $f$ is continuous at $c$ if $\lim\limits_{\large x\to c} f(x) = f(c)$.
Step 1:
At $x=2$
LHL=$\lim\limits_{\large x\to 2}(5)=5$
$f(2)=5$
RHL=$\lim\limits_{\large x\to 2}(ax+b)=2a+b$
$f$ is continuous at $x=2$ if $2a+b=5$
$2a+b=5$----(1)
Step 2:
At $x=10$
LHL=$\lim\limits_{\large x\to 10}f(x)$
$\quad\;=\lim\limits_{\large x\to {10}}(ax+b)$
$\quad\;=10a+b$
RHL=$\lim\limits_{\large x\to 10}f(x)$
$\quad\;=\lim\limits_{\large x\to {10}}(21)$
$\quad\;=21$
Step 3:
$f$ is continuous at $x=10$ if $10a+b=21$
$10a+b=21$------(2)
Subtracting (1) from (2)
$8a=21-5$
$\;\;\;\;\;=16$
$8a=16$
$a=2$
From (1)$\Rightarrow 2\times 2+b=5$
$\qquad\qquad 4+b=5$
$\qquad\qquad b=1$
Therefore $a=2,b=1.$