Chapter 6.4, Problem 8CYU

To determine

To find: the consecutive value of x where the zeroes of the function is located and graph the function.

Answer to Problem 8CYU

The zeroes of the function lied on $\left(-2,-1\right)\cup \left(1,2\right)$ .

Explanation of Solution

Given:

  $\text{f}\left(\text{x}\right){\text{=2x}}^{\text{4}}{\text{-x}}^{\text{3}}{\text{-3x}}^{\text{2}}\text{+2x-4}$ .

Concept used:

The zeroes of the function resembled the function equates to zero.

  $\text{y=f}\left(\text{x}\right)\text{=0}$ .

The zero normally lies in between negative number and positive number.

Therefore, the consecutive number will negative approaches zero and positive approaches to zero except $\text{y=0}$ itself.

The graph of the function $\text{f}$ is the set of all point in the plane of the form $\left(\text{x,f}\left(\text{x}\right)\right)$ .

The graph can be defined by the graph of $\text{f}$ and the equation for the graph will be $\text{y=f}\left(\text{x}\right)$ .

The graph of the function is special case of the graph of an equation.

Calculation:

The function is given as:

  $\text{f}\left(\text{x}\right){\text{=2x}}^{\text{4}}{\text{-x}}^{\text{3}}{\text{-3x}}^{\text{2}}\text{+2x-4}$ .

The table for the graph of the function will be:

By putting the value $\text{x}$ the value of $\text{y}$ can be determine.

$\text{x}$	$-5$	$-3$	$\text{-2}$	$\text{-1}$	$\text{0}$	$\text{1}$	$\text{2}$	$3$
$\text{f}\left(\text{x}\right)$	$1286$	$152$	$\text{20}$	$-6$	$-4$	$-4$	$12$	$110$

The points in the graph so collected will form a graph for the given equation.

The changes in sign from the result indicates that there are zeroes between $\text{x=-2}$ and $\text{x=-1}$ ,and between $\text{x=1}$ and $\text{x=2}$ .

Since, the sign of the function changes from negative to positive.

The graph of the given equation by collecting the points will be:

  

From the graph it easily noticed that graph changes it sign from $\text{x=-2}$ to $\text{x=-1}$ and between $\text{x=1}$ and $\text{x=2}$ .

Hence, the zeroes of the function lied on $\left(-2,-1\right)\cup \left(1,2\right)$ .