Chapter 3.2, Problem 39E

(a)

To determine

Todetermine:The relationship between $a$ and $b$ .

Answer to Problem 39E

The required relationship between $a$ and $b$ is $a+b=2$ .

Explanation of Solution

Given information:

The given function $f\left(x\right)=\{\begin{array}{c}3-x,x<1\\ a{x}^2+bx,x\ge 1\end{array}$

Formula Used:

Condition for continuity:

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

Calculation:

The given function is $f\left(x\right)=\{\begin{array}{c}3-x,x<1\\ a{x}^2+bx,x\ge 1\end{array}$

For $x<1$ , the function is $f\left(x\right)=3-x$

Now, find the continuity for $x<1$ .

  $\begin{array}{c}\underset{x\to 1^{-}}{\lim }f\left(x\right)=f\left(1\right)\\ \underset{x\to 1^{+}}{\lim }f\left(x\right)=f\left(1\right)\\ \underset{x\to 1^{-}}{\lim }f\left(x\right)=\underset{x\to 1^{+}}{\lim }f\left(x\right)\end{array}$

Consider $x\ge 1$ , the function will be $f\left(x\right)=a{x}^2+bx$ .

Find the left hand limit,

  $\begin{array}{c}\underset{x\to 1^{-}}{\lim }f\left(x\right)=\underset{x\to 1^{-}}{\lim }\left[3-x\right]\\ =2\end{array}$

Now, find the right hand limit as

  $\begin{array}{c}\underset{x\to 1^{+}}{\lim }f\left(x\right)=\underset{x\to 1^{+}}{\lim }\left[a{x}^2+bx\right]\\ =a+b\end{array}$

From the condition of continuity

  $\begin{array}{c}\underset{x\to 1^{-}}{\lim }f\left(x\right)=\underset{x\to 1^{+}}{\lim }f\left(x\right)\\ 2=a+b\end{array}$

Hence, the required relationship between $a$ and $b$ is $a+b=2$ .

(b)

To determine

To calculate:The unique values for $a\text{and}b$ so that $f$ is continuous and differentiable.

Answer to Problem 39E

The required values are $a=-3,b=5$ .

Explanation of Solution

Given information:

The given function $f\left(x\right)=\{\begin{array}{c}3-x,x<1\\ a{x}^2+bx,x\ge 1\end{array}$

Formula Used:

Definition of the derivative:

  $\frac{d}{dx}f\left(x\right)=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}$

Where $f\left(x\right)$ is the function and $h$ is a constant.

Calculation:

From 39a $f$ is continuous for all $x$ if $a+b=2$ .

Let $f$ is differentiable at all $x<1$ and $x<1$ .

The right hand derivative is found as

  $\begin{array}{c}\underset{x\to 1^{+}}{\lim }\frac{f\left(x\right)-f\left(1\right)}{x-1}=\underset{x\to 1^{+}}{\lim }\frac{a{x}^2+bx-\left(a+b\right)}{x-1}\\ =\underset{x\to 1^{+}}{\lim }\frac{a{x}^2+bx-a-b}{x-1}\end{array}$

As $a+b=2$ , substitute $b=2-a$ in the expression above:

  $\begin{array}{l}=\underset{x\to 1^{+}}{\lim }\frac{a{x}^2+\left(2-a\right)x-a-\left(2-a\right)}{x-1}\\ =\underset{x\to 1^{+}}{\lim }\frac{a{x}^2+2x-ax-a-2+a}{x-1}\\ =\underset{x\to 1^{+}}{\lim }\frac{a{x}^2+\left(2-a\right)x-2}{x-1}\end{array}$

For the limit above to exist, the numerator must have $x-1$ as a factor.

Now, divide $a{x}^2+\left(2-a\right)x-2$ with $x-1$ .

  $\begin{array}{l}x-1\underset{\_}{\begin{array}{c}\hfill ax+2\\ \hfill \overline{)\begin{array}{l}a{x}^2+\left(2-1a\right)x-2\\ -a{x}^2+ax\end{array}}\end{array}}\\ \underset{\_}{\begin{array}{l}2x-2\\ -2x+2\end{array}}\\ 0\end{array}$

Since, the remainder is zero. Thus, take the limit of $ax+2$ .

  $\begin{array}{l}=\underset{x\to 1^{+}}{\lim }\left[ax+2\right]\\ =a+2\end{array}$

For the differentiability left and right hand derivatives to be equal. Thus,

  $\begin{array}{c}a+2=-1\\ a=-3\end{array}$

Now, calculate the value of $b$ .

  $\begin{array}{l}b=2-a\\ b=5\end{array}$

Hence, the required values are $a=-3,b=5$ .