Chapter 2.1, Problem 43E

a.

To determine

To find: Any two numbers between $1$ to $10$ .

Answer to Problem 43E

The two numbers are $2$ and $3$ .

Explanation of Solution

Given information:

The numbers should be between $1$ to $10$ .

Calculation:

The two numbers are 2 and $3$ .

b.

To determine

To plot: The graph of $\left(x-a\right)\left(x+b\right)$

Explanation of Solution

Given information:

The two numbers be $2$ and $3$ .

Calculation:

Substitute $2$ for $a$ and $3$ by $b$ in $\left(x-a\right)\left(x+b\right)$ .

The function is $\left(x-2\right)\left(x+3\right)$ .

Graph of the function can be plotted as shown below.

  

Conclusion: The graph has been plotted.

c.

To determine

To plot: The graph of the derivative of the function.

Explanation of Solution

Given information:

The function is $y=x^2-5x+6$ and the slope of the line is $2$ .

Concept used:

The derivative of the function is the slope of the line and the standard equation of line is $y=mx+c$ .

Calculation:

The function is $y=x^2-5x+6$ .

Differentiate the function with respect to $x$ .

  $\begin{array}{c}\frac{dy}{dx}=\frac{d}{dx}\left(x^2-5x+6\right)\\ =\frac{d}{dx}\left(x^2\right)-5\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(6\right)\\ m=2x-5\end{array}$

Substitute $2$ for $m$ in the line $m=2x-5$ .

  $\begin{array}{c}2=2x-5\\ 2+5=2x\\ x=\frac{7}{2}\end{array}$

Conclusion: The value of $\underset{x\to 1^{-}}{\lim }{f}^{\prime }\left(x\right)=2$

d.

To determine

To find: The value of $\underset{x\to 1^{+}}{\lim }{f}^{\prime }\left(x\right)$

Answer to Problem 43E

The value is $2$ .

Explanation of Solution

Given information:

The function is $f\left(x\right)=\left\{\begin{array}{c}{x}^2,x\le 1\\ 2x,x>1\end{array}\right.$

Concept used:

The right-hand derivative is given by the formula $f^{\prime }\left(x\right)=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}$ .

Calculation:

Here, in the piece wise function the value of $f\left(x\right)$ at $x<1$ is $2x$ .

Substitute the values in the formula,

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

Substitute the values in the formula,

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

Conclusion: The value of $f^{\prime }\left(x\right)=2$

e.

To determine

Existence of $\underset{x\to 1}{\lim }{f}^{\prime }\left(x\right)$ .

Answer to Problem 43E

The $\underset{x\to 1}{\lim }{f}^{\prime }\left(x\right)$ exists.

Explanation of Solution

Given information:

The function is $f\left(x\right)=\left\{\begin{array}{c}{x}^2,x\le 1\\ 2x,x>1\end{array}\right.$

Concept used:

If $\underset{x\to 1^{-}}{\lim }{f}^{\prime }\left(x\right)=\underset{x\to 1^{+}}{\lim }{f}^{\prime }\left(x\right)$ , then $\underset{x\to 1}{\lim }{f}^{\prime }\left(x\right)$ exists.

Calculation:

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

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

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

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

Conclusion: The value of $\underset{x\to 1}{\lim }{f}^{\prime }\left(x\right)$ exists.