Chapter 2, Problem 33CR

(a)

To determine

To graph: Applying leading coefficient test to the function and draw the graph.

Explanation of Solution

Given information: 

  $f(x)=x^4-2x^3-12x+18x+27$

Calculation:

By leading coefficient test:

As the x⁴is a leading variable and its coefficient is 1.

So when the large numbers were taken than x⁴ will be dominating

For $x<0$ negative large values are taken and the answer will be a positive number so the graph increases for $x<0$

For $x>0$ positive large values are taken and the answer will be a positive number so the graph increases for $x>0$

The graph will be

  

(b)

To determine

To calculate: Zeros of the polynomial.

Answer to Problem 33CR

The curve will pass through $x=-3,-1,3$

Explanation of Solution

Given information: 

  $f(x)=x^4-2x^3-12x+18x+27$

Calculation: 

Zeros of the polynomial are:

  $\begin{array}{l}\text{Use the rational root theorem}\\ {\text{a}}_{\text{0}}{\text{=27,a}}_{\text{n}}\text{=1}\\ \text{(x}+\text{1)}\frac{{\text{x}}^{\text{4}}-2x^3-12x^2+18x+27}{x+1}=0\\ \text{(x}+\text{1)(}{x}^3-3x^2-9x+27)=0\\ \text{(x}+\text{1)[}{x}^2(x-3)-9(x-3)]=0\\ \text{(x}+\text{1)}(x^2-9)(x-3)=0\\ $ {(x-3)}^2(x+3)(x+1)=0$\end{array}$

  $x=-3,-1,3,3$

The curve will pass through $x=-3,-1,3$

(c)

To determine

To graph: points on the graph.

Answer to Problem 33CR

So the curve also passes through (-1,0) (0,27) (1,32)

Explanation of Solution

Given information: 

  ${(x-3)}^2(x+3)(x+1)$

Calculation: 

Plotting points:

  ${(x-3)}^2(x+3)(x+1)$

F(x)	x
0	-1
27	0
32	1

So the curve also passes through (-1, 0), (0, 27), (1, 32)

Points plotted on the graph are given below:

  

(d)

To determine

To graph:  $f(x)=x^4-2x^3-12x+18x+27$

Explanation of Solution

Given information: 

  $f(x)=x^4-2x^3-12x+18x+27$

Graph: 

  

Interpretation: It is a continuous curve passes through (-1, 0), (0, 27), (1, 32).