Question

The function $f$ is defined as $f\left(x\right)=\{\begin{array}{cc}{x}^2+ax+b,& \text{if}0\le x<2\\ 3x+2,& \text{if}2\le x\le 4\\ 2ax+5b,& \text{if}4<x\le 8\end{array}$ , If $f$ is continuous in $[0,8]$ find the values of $a$ and $b$.

Answer 1

Given,

$f\left(x\right)=\{\begin{array}{cc}{x}^2+ax+b,& \text{if}0\le x\le 2\\ 3x+2,& \text{if}2\le x\le 4\\ 2ax+5b,& \text{if}4\le x\le 8\end{array}$

Given, $f\left(x\right)$ is continuous in $[0,8]$.

$\therefore$ $f\left(x\right)$ is continuous at $x=2$ and $x=4$.

At $x=2$, we have

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

$\underset{x\to 2^{+}}{lim}f\left(x\right)=\underset{h\to 0}{lim}f(2+h)=\underset{h\to 0}{lim}[3(2+h)+2]=8$

$\therefore$ $\underset{x\to 2^{-}}{lim}f\left(x\right)=\underset{x\to 2^{+}}{lim}f\left(x\right)$

$\Rightarrow 4+2a+b=8$

$\Rightarrow 2a+b=4.....\left(1\right)$

Also, at $x=4$

$\underset{x\to 4^{-}}{lim}f\left(x\right)=\underset{h\to 0}{lim}f(4-h)=\underset{h\to 0}{lim}[3(4-h)+2]=14$

$\underset{x\to 4^{+}}{lim}f\left(x\right)=\underset{h\to 0}{lim}f(4+h)=\underset{h\to 0}{lim}[2a(4+h)+5b]=8a+5b$

$\therefore$ $\underset{x\to 4^{+}}{lim}f\left(x\right)=\underset{x\to 4^{-}}{lim}f\left(x\right)$

$\therefore$ $8a+5b=14.....\left(2\right)$

Solving equation $\left(1\right)$ and $\left(2\right)$, we get

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