The function f(x) is defined as follows:

Solution:

Given: $f$ is continuous on $[0,8]$.

$\therefore f$ is continuous at $x=2$ and $x=4$

At $x=2$, we have

$\lim _{x \rightarrow 2^{-}} f(x)=\lim _{h \rightarrow 0} f(2-h)=\lim _{h \rightarrow 0}\left[(2-h)^{2}+a(2-h)+b\right]=4+2 a+b$

$\lim _{x \rightarrow 2^{+}} f(x)=\lim _{h \rightarrow 0} f(2+h)=\lim _{h \rightarrow 0}[3(2+h)+2]=8$

Also,

At $x=4$, we have

$\lim _{x \rightarrow 4^{-}} f(x)=\lim _{h \rightarrow 0} f(4-h)=\lim _{h \rightarrow 0}[3(4-h)+2]=14$

$\lim _{x \rightarrow 4^{+}} f(x)=\lim _{h \rightarrow 0} f(4+h)=\lim _{h \rightarrow 0}[2 a(4+h)+5 b]=8 a+5 b$

$f$ is continuous at $x=2$ and $x=4$

$\therefore \lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2^{+}} f(x)$ and $\lim _{x \rightarrow 4^{-}} f(x)=\lim _{x \rightarrow 4^{+}} f(x)$

$\Rightarrow 4+2 a+b=8$ and $8 a+5 b=14$

$\Rightarrow 2 a+b=4 \quad \ldots(1)$ and $8 a+5 b=14 \quad \ldots(2)$

On simplifying eqs. (1) and (2), we get

$a=3$ and $b=-2$