Chapter 6.4, Problem 46PPS

(a)

To determine

To Find: The degree, leading coefficient and the end behaviour.

Answer to Problem 46PPS

The degree is 1, the leading coefficient is 1 and the end behaviour is that both ends go up.

Explanation of Solution

Given:

The given function is $f\left(x\right)=x^4-8.65x^3+27.34x^4-37.2285x+18.27$ .

Calculation:

From the given function the degree is 4 as it is highest power of the function.

The leading coefficient 1 as it is the coefficient of the variable of the highest degree.

Consider the end behaviours as it is even degree polynomial and the leading coefficient is positive so the ends that will both go up.

(b)

To determine

To Find:The table of the integer values $f\left(x\right)$ if $-4\le x\le 4$ . How many zeroes does the function appear to have from the table.

Answer to Problem 46PPS

Therequired table for the 10 zeroes is shown in Table 1

Explanation of Solution

Consider the given function is,

  $f\left(x\right)=x^4-8.65x^3+27.34x^4-37.2285x+18.27$

The table to the integer values $\left(x\in \left[-4,4\right]\right)$ is shown in Table 1

Table 1

$x$	$f\left(x\right)$
-4	1414.224
-3	690.5655
-2	287.287
-1	92.4885
0	18.27
1	0.7315
2	-0.027
3	0.0945
4	9.916

The above table shows two change of signs and the number of zeroes.

Thus, the required table for the 10 zeroes is shown in Table 1

(c)

To determine

To Find:The graph for the function.

Answer to Problem 46PPS

The required graph is shown in Figure 1 and the zeroes from the graph is $x_{1,2,3,4}\approx 1.25,2.1,2.4,2.9$ .

Explanation of Solution

Consider the given function is,

  $f\left(x\right)=x^4-8.65x^3+27.34x^4-37.2285x+18.27$

The graph for the above function is shown in Figure 1

  

Figure 1

(d)

To determine

To Find:The change in the viewing window to $\left[0,4\right]\text{scl}$ by $\left[-0.4,0.4\right]\text{scl:0}\text{.2}$ . The conclusion that can be made from the new view of the graph.

Answer to Problem 46PPS

The given graph is shown in Figure 2

Explanation of Solution

From the given graph shown in Figure the graph for the change in viewing window with the points marked is shown in Figure 2

  

Figure 2

From the above graph from the interval 1 to 3 there are four zeroes but not only two as it is determined in part (b) as,

  $\begin{array}{l}{x}_1\approx 1.25\\ x_2\approx 2.1\\ x_3\approx 2.4\\ x_4\approx 2.9\end{array}$