Chapter 2.2, Problem 39E

a.

To determine

To find: The relation between $a$ , $b$ .

Answer to Problem 39E

The relation is $a+b=2$ .

Explanation of Solution

Given information:

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

Concept used:

If $\underset{x\to a}{\lim }f\left(x\right)=\underset{x\to a^{-}}{\lim }f\left(x\right)=f\left(a\right)$ then the function is said to be continuous.

Calculation:

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

Value of limit $\underset{x\to 1^{+}}{\lim }f\left(x\right)$ is as shown below.

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

Calculation of value of $\underset{x\to 1^{-}}{\lim }f\left(x\right)$ is as shown below.

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

Calculation of value of $\underset{x\to 1}{\lim }f\left(x\right)$ is as shown below.

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

Since, the function is continuous, equate to find the limits.

  $a+b=2$

Conclusion: The relation is $a+b=2$ .

b.

To determine

To find: The values are $a$ , $b$ .

Answer to Problem 39E

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

Explanation of Solution

Given information:

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

Concept used:

If $\underset{x\to a}{\lim }f\left(x\right)=\underset{x\to a^{-}}{\lim }f\left(x\right)=f\left(a\right)$ then the function is said to be continuous, the right hand derivative of a function is $\underset{h\to 0^{+}}{\lim }\frac{f\left(1+h\right)-f\left(1\right)}{h}$ and the left hand derivative of a function is $\underset{h\to 0^{-}}{\lim }\frac{f\left(1+h\right)-f\left(1\right)}{h}$

Calculation:

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

Value of limit $\underset{x\to 1^{+}}{\lim }f\left(x\right)$ is as shown below.

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

Calculation of value of $\underset{x\to 1^{-}}{\lim }f\left(x\right)$ is as shown below.

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

Calculation of value of $\underset{x\to 1}{\lim }f\left(x\right)$ is as shown below.

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

Since the function is continuous, so the left hand limit and right hand limit are equal.

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

Value of limit $\underset{h\to 0^{-}}{\lim }\frac{f\left(1+h\right)-f\left(1\right)}{h}$ is as shown below.

  $\begin{array}{c}\underset{h\to 0^{-}}{\lim }\frac{f\left(1+h\right)-f\left(1\right)}{h}=\underset{h\to 0^{-}}{\lim }\frac{\left(3-\left(1+h\right)\right)-\left(a{\left(1\right)}^2+b\left(1\right)\right)}{h}\\ =\underset{h\to 0^{-}}{\lim }\frac{3-1-h-a-b}{h}\\ =\underset{h\to 0^{-}}{\lim }\frac{2-h-a-b}{h}\\ =-1\end{array}$

Value of limit $\underset{h\to 0^{+}}{\lim }\frac{f\left(1+h\right)-f\left(1\right)}{h}$ is as shown below.

  $\begin{array}{c}\underset{h\to 0^{+}}{\lim }\frac{f\left(1+h\right)-f\left(1\right)}{h}=\underset{h\to 0^{+}}{\lim }\frac{\left(a{\left(1+h\right)}^2+b\left(1+h\right)\right)-\left(a{\left(1\right)}^2+b\left(1\right)\right)}{h}\\ =\underset{h\to 0^{+}}{\lim }\frac{a+a{h}^2+2ah+b+bh-a-b}{h}\\ =\underset{h\to 0^{+}}{\lim }\frac{h\left(ah+2a+b\right)}{h}\\ =2a+b\end{array}$

To make the function differentiable, the right hand and left hand derivative are equal.

  $2a+b=-1$ .

Substitute $2-b$ for $a$ in $2a+b=-1$ .

  $\begin{array}{c}2\left(2-b\right)+b=-1\\ 4-2b+b=-1\\ 4-b=-1\\ -b=-1-4\\ b=5\end{array}$

Substitute $5$ for $b$ in the equation $2a+b=-1$ .

  $\begin{array}{c}2a+5=-1\\ 2a=-6\\ a=-\frac{6}{2}\\ a=-3\end{array}$

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