Chapter 3.5, Problem 39E

To determine

To find: the values of $a$and $b$ to make f continuous.

Answer to Problem 39E

  $\begin{array}{l}a=4\\ b=2\end{array}$

Explanation of Solution

Given information:

  $f\left(x\right)=\{\begin{array}{l}{x}^2+a\text{ if }x\ge 2\\ bx+a\text{ if }-2<x<2\\ \sqrt{-b-x}\text{ if }x\le -2\end{array}$

Calculation:

In order to make the function continuous at $x=2$ , the graph needs to be such that first two functions meet at $x=2$ . Thus, set both the functions to equal and solve as follows:

  $\begin{array}{c}{x}^2+a=bx+a\\ 2^2+a=b\left(2\right)+a\left[\text{Put }x=2\right]\\ 4+a=2b+a\\ b=\frac{4}{2}\\ =2\end{array}$

The second function must meet at $x=-2$ in order to maintain the continuity. So, set them equal and solve as follows:

  $\begin{array}{c}2x+4=\sqrt{-2-x}\\ 2\left(-2\right)+a=\sqrt{-2-\left(-2\right)}\left[\text{Put }x=-2\text{ and }b=2\right]\\ -4+a=\sqrt{-2+2}\\ -4+a=0\\ a=4\end{array}$

Thus, the value of a and b are 4 and 2 respectively to make the given function continuous.