Chapter 2.2, Problem 47E

a.

To determine

To find: The graph of piecewise function $f\left(x\right)=\left\{\begin{array}{l}{x}^2,x\le 0\\ 2x,x>0\end{array}\right.$

Explanation of Solution

Given:

The piece wise function is given as:

  $f\left(x\right)=\left\{\begin{array}{l}{x}^2,x\le 0\\ 2x,x>0\end{array}\right.$

Window is given $\text{[-4}\text{.7,4}\text{.7] }$ and $\text{[-3,5]}$ .

Calculation:

The graph of the piecewise function $f\left(x\right)$ is given below:

  

The graph of piece wise function $Y1$ in the given window is below:

  

b.

To determine

To find: Explain why the value of $f\left(x\right)$ and $Y1$ is same.

Answer to Problem 47E

Explanation given.

Explanation of Solution

Given:

The piece wise function is given as:

  $f\left(x\right)=\left\{\begin{array}{l}{x}^2,x\le 0\\ 2x,x>0\end{array}\right.$

Both the graphs are same because both functions are equal the graphing notation of $f\left(x\right)$ in graphing calculator is $Y1$ also $Y1$ is restricted in the window $\text{[-4}\text{.7,4}\text{.7] }$ and $\text{[-3,5]}$ .

This only makes graph more clear but the graph is same as the function is same for both graphs.

c.

To determine

To find: Find the numerical derivative of $Y1$ and graph it.

Answer to Problem 47E

The numerical derivatives of $Y1$ are $2x,2$ respectively.

Explanation of Solution

Given:

The piece wise function is given as:

  $f\left(x\right)=\left\{\begin{array}{l}{x}^2,x\le 0\\ 2x,x>0\end{array}\right.$

Concept used: Numerical derivative is defined as NDER $\left(f\left(x\right),x\right)=\frac{f\left(x+0.001\right)-f\left(x-0.001\right)}{0.002}$

Calculation:

Numerical derivative is defined as NDER $\left(f\left(x\right),x\right)=\frac{f\left(x+0.001\right)-f\left(x-0.001\right)}{0.002}$

Find $f\left(x+0.001\right)$ by substituting $x=x+0.001$ in $f\left(x\right)=x^2$ .

  $\begin{array}{l}f\left(x+0.001\right)={\left(x+0.001\right)}^2\\ f\left(x+0.001\right)=x^2+0.002x+{\left(0.001\right)}^2\end{array}$

Find $f\left(x-0.001\right)$ by substituting $x=x-0.001$ in $f\left(x\right)=x^2$ .

  $\begin{array}{l}f\left(x-0.001\right)={\left(x-0.001\right)}^2\\ f\left(x+0.001\right)=x^2-0.002x+{\left(0.001\right)}^2\end{array}$

Now substitute the values of $f\left(x+0.001\right)$ and $f\left(x-0.001\right)$ in NDER $\left(f\left(x\right),x\right)$ :

  $\begin{array}{l}\text{NDER}\left(\text{f}\left(\text{x}\right),x\right)=\frac{x^2+0.002x+{\left(0.001\right)}^2-\left(x^2-0.002x+{\left(0.001\right)}^2\right)}{0.002}\\ \text{NDER}\left(\text{f}\left(\text{x}\right),x\right)=\frac{x^2+0.002x+{\left(0.001\right)}^2-x^2+0.002x-{\left(0.001\right)}^2}{0.002}\\ \text{NDER}\left(\text{f}\left(\text{x}\right),x\right)=\frac{0.004x}{0.002}\\ \text{NDER}\left(\text{f}\left(\text{x}\right),x\right)=2x\end{array}$

Find $f\left(x+0.001\right)$ by substituting $x=x+0.001$ in $f\left(x\right)=2x$ .

  $\begin{array}{l}f\left(x+0.001\right)=2\left(x+0.001\right)\\ f\left(x+0.001\right)=2x+0.002\end{array}$

Find $f\left(x-0.001\right)$ by substituting $x=x-0.001$ in $f\left(x\right)=x^2$ .

  $\begin{array}{l}f\left(x-0.001\right)=2\left(x-0.001\right)\\ f\left(x-0.001\right)=2x-0.002\end{array}$

Now substitute the values of $f\left(x+0.001\right)$ and $f\left(x-0.001\right)$ in NDER $f\left(x\right),x$:

  $\begin{array}{l}\text{NDER}\left(\text{f}\left(\text{x}\right),x\right)=\frac{2x+0.002-\left(2x-0.002\right)}{0.002}\\ \text{NDER}\left(\text{f}\left(\text{x}\right),x\right)=\frac{2x+0.002-2x+0.002}{0.002}\\ \text{NDER}\left(\text{f}\left(\text{x}\right),x\right)=\frac{0.004}{0.002}\\ \text{NDER}\left(\text{f}\left(\text{x}\right),x\right)=2\end{array}$

The graph of $Y2$ is given below:

  

d.

To determine

To find: NDER $\left(Y1,-0.1\right),\left(Y1,-0\right)\left(Y1,0.1\right)$ and graph it.

Answer to Problem 47E

The value of NDER $\left(Y1,-0.1\right),\left(Y1,-0\right)\left(Y1,0.1\right)$ are $\left(-0.2,2\right)\left(0,2\right)$ and $\left(0.2,2\right)$ respectively.

Explanation of Solution

Given:

The piece wise function is given as:

  $f\left(x\right)=\left\{\begin{array}{l}{x}^2,x\le 0\\ 2x,x>0\end{array}\right.$

Concept used: Numerical derivative is defined as NDER $\left(f\left(x\right),x\right)=\frac{f\left(x+0.001\right)-f\left(x-0.001\right)}{0.002}$

Calculation:

Numerical derivative is defined as NDER $\left(Y1,-0.1\right)=\frac{f\left(-0.1+0.001\right)-f\left(-0.1-0.001\right)}{0.002}$

Find $f\left(-0.1+0.001\right)$ by substituting $x=-0.1+0.001$ in $f\left(x\right)=x^2$ .

  $\begin{array}{l}f\left(-0.1+0.001\right)={\left(-0.099\right)}^2\\ f\left(-0.099\right)={\left(-0.099\right)}^2\end{array}$

Find $f\left(-0.1-0.001\right)$ by substituting $x=-0.1-0.001$ in $f\left(x\right)=x^2$ .

  $\begin{array}{l}f\left(-0.1-0.001\right)={\left(-.101\right)}^2\\ f\left(-0.101\right)={\left(-.101\right)}^2\end{array}$

Now substitute the values of $f\left(-0.099\right)$ and $f\left(-0.101\right)$ in NDER $\left(Y1,-0.1\right)$ :

  $\begin{array}{l}\text{NDER}\left(\text{Y1, }\text{-0}\text{.1}\right)=\frac{{\left(-0.099\right)}^2-{\left(-0.101\right)}^2}{0.002}\\ \text{NDER}\left(\text{Y1, }\text{-0}\text{.1}\right)=\frac{-0.0004}{0.002}\\ \text{NDER}\left(\text{Y1, }\text{-0}\text{.1}\right)=-0.2\end{array}$

Find $f\left(-0.1+0.001\right)$ by substituting $x=-0.1+0.001$ in $f\left(x\right)=2x$ .

  $\begin{array}{l}f\left(-0.1+0.001\right)=2\left(-0.099\right)\\ f\left(-0.099\right)=-0.198\end{array}$

Find $f\left(-0.1-0.001\right)$ by substituting $x=-0.1-0.001$ in $f\left(x\right)=2x$ .

  $\begin{array}{l}f\left(-0.1-0.001\right)=2\left(-.1001\right)\\ f\left(-0.101\right)=-0.202\end{array}$

Now substitute the values of $f\left(-0.099\right)$ and $f\left(-0.1001\right)$ in NDER $\left(Y1,-0.1\right)$ :

  $\begin{array}{l}\text{NDER}\left(\text{Y1, }\text{-0}\text{.1}\right)=\frac{-0.198-\left(-0.202\right)}{0.002}\\ \text{NDER}\left(\text{Y1, }\text{-0}\text{.1}\right)=\frac{-0.198+0.202}{0.002}\\ \text{NDER}\left(\text{Y1, }\text{-0}\text{.1}\right)=\frac{0.004}{.002}\\ \text{NDER}\left(\text{Y1, }\text{-0}\text{.1}\right)=2\end{array}$

Numerical differentiation NDER $\left(Y1,-0.1\right)$ of $f\left(x\right)=x^2$ is -0.2.

Numerical differentiation NDER $\left(Y1,-0.1\right)$ of $f\left(x\right)=2x$ is 2.

Numerical derivative is defined as NDER $\left(Y1,0\right)=\frac{f\left(0+0.001\right)-f\left(0-0.001\right)}{0.002}$ .

Find $f\left(0+0.001\right)$ by substituting $x=0+0.001$ in $f\left(x\right)=x^2$ .

  $\begin{array}{l}f\left(0+0.001\right)={\left(0.001\right)}^2\\ f\left(0.001\right)=0.0000001\end{array}$

Find $f\left(0-0.001\right)$ by substituting $x=0-0.001$ in $f\left(x\right)=x^2$ .

  $\begin{array}{l}f\left(0-0.001\right)={\left(-.001\right)}^2\\ f\left(-0.001\right)=0.0000001\end{array}$

Now substitute the values of $f\left(0.001\right)$ and $f\left(-0.001\right)$ in NDER $\left(Y1,0\right)$ :

  $\begin{array}{l}\text{NDER}\left(\text{Y1, }\text{0}\right)=\frac{0.0000001-0.0000006}{0.002}\\ \text{NDER}\left(\text{Y1, }\text{0}\right)=\frac{0}{0.002}\\ \text{NDER}\left(\text{Y1, }\text{0}\right)=0\end{array}$

Find $f\left(0+0.001\right)$ by substituting $x=0+0.001$ in $f\left(x\right)=2x$ .

  $\begin{array}{l}f\left(0.001\right)=2\left(0.001\right)\\ f\left(0.001\right)=0.002\end{array}$

Find $f\left(0-0.001\right)$ by substituting $x=0-0.001$ in $f\left(x\right)=2x$ .

  $\begin{array}{l}f\left(-0.001\right)=2\left(-.001\right)\\ f\left(-0.001\right)=-0.002\end{array}$

Now substitute the values of $f\left(0.001\right)$ and $f\left(-0.001\right)$ in NDER $\left(Y1,0\right)$ :

  $\begin{array}{l}\text{NDER}\left(\text{Y1, }\text{0}\right)=\frac{0.002-\left(-0.002\right)}{0.002}\\ \text{NDER}\left(\text{Y1, }\text{0}\right)=\frac{0.002+0.002}{0.002}\\ \text{NDER}\left(\text{Y1, }\text{0}\right)=\frac{0.004}{.002}\\ \text{NDER}\left(\text{Y1, }\text{0}\right)=2\end{array}$

Numerical differentiation NDER $\left(Y1,0\right)$ of $f\left(x\right)=x^2$ is 0.

Numerical differentiation NDER $\left(Y1,0\right)$ of $f\left(x\right)=2x$ is 2.

Numerical derivative is defined as NDER $\left(Y1,0.1\right)=\frac{f\left(0.1+0.001\right)-f\left(0.1-0.001\right)}{0.002}$ .

Find $f\left(0.1+0.001\right)$ by substituting $x=0.1+0.001$ in $f\left(x\right)=x^2$ .

  $\begin{array}{l}f\left(0.1+0.001\right)={\left(0.101\right)}^2\\ f\left(0.101\right)=.010201\end{array}$

Find $f\left(0.1-0.001\right)$ by substituting $x=0.1-0.001$ in $f\left(x\right)=x^2$ .

  $\begin{array}{l}f\left(0.1-0.001\right)={\left(0.099\right)}^2\\ f\left(0.099\right)=0.009801\end{array}$

Now substitute the values of $f\left(0.101\right)$ and $f\left(0.099\right)$ in NDER $\left(Y1,0.1\right)$ :

  $\begin{array}{l}\text{NDER}\left(\text{Y1, }\text{0}\text{.1}\right)=\frac{{\left(.101\right)}^2-{\left(-0.099\right)}^2}{0.002}\\ \text{NDER}\left(\text{Y1, }\text{0}\text{.1}\right)=\frac{0.010201-0.009801}{0.002}\\ \text{NDER}\left(\text{Y1, }\text{0}\text{.1}\right)=\frac{0.0004}{0.002}\\ \text{NDER}\left(\text{Y1, }\text{0}\text{.1}\right)=0.2\end{array}$

Find $f\left(0.1+0.001\right)$ by substituting $x=0.1+0.001$ in $f\left(x\right)=2x$ .

  $\begin{array}{l}f\left(0.1+0.001\right)=2\left(0.101\right)\\ f\left(0.101\right)=0.202\end{array}$

Find $f\left(0.1-0.001\right)$ by substituting $x=0.1-0.001$ in $f\left(x\right)=2x$ .

  $\begin{array}{l}f\left(0.1-0.001\right)=2\left(.099\right)\\ f\left(0.099\right)=0.198\end{array}$

Now substitute the values of $f\left(0.101\right)$ and $f\left(0.099\right)$ in NDER $\left(Y1,0.1\right)$ :

  $\begin{array}{l}\text{NDER}\left(\text{Y1, }\text{0}\text{.1}\right)=\frac{0.202-0.198}{0.002}\\ \text{NDER}\left(\text{Y1, }\text{0}\text{.1}\right)=\frac{0.004}{.002}\\ \text{NDER}\left(\text{Y1, }\text{0}\text{.1}\right)=2\end{array}$

Numerical differentiation NDER $\left(Y1,0.1\right)$ of $f\left(x\right)=x^2$ is 0.2.

Numerical differentiation NDER $\left(Y1,-0.1\right)$ of $f\left(x\right)=2x$ is 2.