Question

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

Answer 1

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

∴ f is continuous at x = 2 and x = 4

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

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

Also,
At x = 4, we have

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

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

f is continuous at x = 2 and x = 4

∴ $\underset{x\to 2^{-}}{\lim }f\left(x\right)=\underset{x\to 2^{+}}{\lim }f\left(x\right)\mathrm{and}\underset{x\to 4^{-}}{\lim }f\left(x\right)=\underset{x\to 4^{+}}{\lim }f\left(x\right)$

$$\begin{gathered} \Rightarrow 4+2a+b=8\mathrm{and}8a+5b=14 \\ \Rightarrow 2a+b=4...\left(1\right)\mathrm{and}8a+5b=14...\left(2\right) \end{gathered}$$

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

$a=3\mathrm{and}b=-2$