Chapter 3.2, Problem 47E

(a)

To determine

To find: The enter the expression $"\left(X^2\right)\left(X\le 0\right)+\left(2X\right)\left(X>0\right)"$ into Y1 of calculator and draw the graph in the window $\left[-4.7,4.7\right]$ by $\left[-3,5\right]$ .

Explanation of Solution

Given:

The given piecewise function for the graph is shown in Figure 1

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

Calculation:

The graph for the given inequalities is shown in Figure 1

  

Figure 1

(b)

To determine

To find: Thereason for the values of Y1 and $f\left(x\right)$ are the same.

Answer to Problem 47E

The value of Y1 is $x^2$ for every value of $x<0$ and for $2x$ and for every $x\ge 0$ and the piecewise function is $f\left(x\right)=\{\begin{array}{l}{x}^2,x\le 0\\ 2x,x>0\end{array}$ .

Explanation of Solution

Use the trace to help see that the value of Y1 is $x^2$ for every value of $x<0$ and for $2x$ and for every $x\ge 0$ and the piecewise function is,

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

Thus, the value of the Y1 and $f\left(x\right)$ are the same.

(c)

To determine

To find: The numerical derivative of Y1 into Y2 of your calculator and draw the graph in the same window and the then turn of the graph Y1.

Explanation of Solution

Draw the graph with calculator as shown in Figure 2

  

Figure 2

(d)

To determine

To find: The trace to calculate NDER $NDER\left(Y1,x,-0.1\right),NDER\left(Y1,x,0\right),NDER\left(Y1,x,0.1\right)$ and compare it with the section 3.1 of Example 6

Answer to Problem 47E

The values are $\text{NDER}\left(Y1,x,-0.1\right)=-0.1$ , $\text{NDER}\left(Y1,x,0\right)=0.9995$ and $\text{NDER}\left(Y1,x,0.1\right)=2$ .

Explanation of Solution

Trace from the given data to determine the value as,

  $\begin{array}{l}\text{NDER}\left(Y1,x,-0.1\right)=-0.1\\ \text{NDER}\left(Y1,x,0\right)=0.9995\\ \text{NDER}\left(Y1,x,0.1\right)=2\end{array}$