a.
To calculate the domain of the function f(x)=6x2−11x+33x2−x .
a.

Answer to Problem 55RE
R−{0,0.333}
Explanation of Solution
Given:
Function: f(x)=6x2−11x+33x2−x
Concept used:
The domain of the function is the set of values, where the function is defined.
Calculation:
Now, finding the domain of the function,
When x=0,x=0.333 , the function is undefined.
So, the domain of the function is R−{0,0.333}.
Conclusion:
Hence, the domain of the function is R−{0,0.333} .
b.
To find the intercepts of the function f(x)=6x2−11x+33x2−x .
b.

Answer to Problem 55RE
(1.5, 0)
Explanation of Solution
Given:
Function: f(x)=6x2−11x+33x2−x
Calculation:
Consider the function f(x)=6x2−11x+33x2−x ,
⇒f(x)=6x2−11x+33x2−x⇒f(x)=6x2−9x−2x+3x(3x−1)⇒f(x)=3x(2x−3)−(2x−3)x(3x−1)⇒f(x)=(2x−3)(3x−1)x(3x−1)⇒f(x)=2x−3x
To find the y-intercept, put x=0 in given function.
⇒f(0)=0−30=−30→Undefined
So, y-intercept does not exist.
To find the x-intercept, put f(x)=y=0 in given function.
⇒0=2x−3x⇒0=2x−3⇒2x=3⇒x=32=1.5
So, the x- intercept is (1.5, 0).
Conclusion:
Therefore, the x- intercept is (1.5, 0).
c.
To find the asymptotes to the function f(x)=6x2−11x+33x2−x .
c.

Answer to Problem 55RE
x=0,y=2
Explanation of Solution
Given:
Function: f(x)=6x2−11x+33x2−x
Calculation:
⇒f(x)=6x2−11x+33x2−x⇒f(x)=6x2−9x−2x+3x(3x−1)⇒f(x)=3x(2x−3)−(2x−3)x(3x−1)⇒f(x)=(2x−3)(3x−1)x(3x−1)⇒f(x)=2x−3x
Asymptotes:
Vertical asymptotes:
To find vertical asymptotes, put denominator of the given function equal to zero.
⇒x=0
Horizontal asymptotes:
As the degree of numerator is equal to the degree of the denominator, the horizontal asymptote for given function is
⇒y=leading coefficient of numeratorleading coefficient of denominator⇒y=21⇒y=2
Conclusion:
Therefore, the vertical asymptote is x=0 and horizontal asymptote is y = 2.
d.
To sketch the function f(x)=6x2−11x+33x2−x .
d.

Explanation of Solution
Given:
Function: f(x)=6x2−11x+33x2−x
Calculation for graph:
Consider f(x)=6x2−11x+33x2−x
Values of x | Values of h(x) |
0 | Infinity |
1 | -1 |
-1 | 5 |
2 | 0.5 |
-2 | 3.5 |
By taking different values of x, the graph can be plotted.
Graph:
Interpretation:
The above graph represents the sketch of given function.
Chapter 2 Solutions
EBK PRECALCULUS W/LIMITS
- can you solve this question using partial fraction decomposition and explain the steps used along the wayarrow_forwardIntegral How 80*1037 IW 1012 S е ऍ dw answer=0 How 70+10 A 80*1037 Ln (Iwl+1) du answer=123.6K 70*1637arrow_forwardcan you solve this question and explain the steps used along the wayarrow_forward
- can you solve this question and explain the steps used along the wayarrow_forwardcan you solve this question and explain the steps used along the wayarrow_forwardCan the expert solve an Intestal In detall? 110x/0³ W. 1 SW = dw A 40x103π ⑤M-1 大 80*10³/ 12 10% 70*1037 80x103 || dw OP= # Sin (w/+1) dw A 70*10*Aarrow_forward
- Calculus: Early TranscendentalsCalculusISBN:9781285741550Author:James StewartPublisher:Cengage LearningThomas' Calculus (14th Edition)CalculusISBN:9780134438986Author:Joel R. Hass, Christopher E. Heil, Maurice D. WeirPublisher:PEARSONCalculus: Early Transcendentals (3rd Edition)CalculusISBN:9780134763644Author:William L. Briggs, Lyle Cochran, Bernard Gillett, Eric SchulzPublisher:PEARSON
- Calculus: Early TranscendentalsCalculusISBN:9781319050740Author:Jon Rogawski, Colin Adams, Robert FranzosaPublisher:W. H. FreemanCalculus: Early Transcendental FunctionsCalculusISBN:9781337552516Author:Ron Larson, Bruce H. EdwardsPublisher:Cengage Learning





