(a)
To state: the domain of the given function.
(a)

Answer to Problem 57E
The domain of the given function is (−∞,1)∪(1,∞) .
Explanation of Solution
Given information:
Consider f(x)=x2−x+1x−1
Calculation:
The domain is the set of x values for which the function is defined.
Find the undefined values of f(x)=x2−x+1x−1
x−1=0x=1
The function is define for all real values of x except at x=1
Hence, the domain is (−∞,1)∪(1,∞)
(b)
To identify: all the intercepts of the given function.
(b)

Answer to Problem 57E
The y- intercept is (0,−1)
Explanation of Solution
Given information:
Consider f(x)=x2−x+1x−1
Calculation:
The x- intercept is the point at which the graph of the function crosses the x- axis.
The y- intercept is the point at which the graph of the function crosses the y- axis.
y=0⇒y=x2−x+1x−1=0x is not a real
x=0⇒y=f(0)=0+10−1=−1
Hence, y- intercept is (0,−1)
(c)
To find: the vertical or slant asymptotes of the given function.
(c)

Answer to Problem 57E
The oblique asymptote is y=x ,vertical asymptote x=1
Explanation of Solution
Given information:
Consider f(x)=x2−x+1x−1
Calculation:
Degree of numerator is greater than the degree of denominator. So there is no horizontal
Asymptote. Slant asymptote is there.
y=x2−x+1x−1=x+1x−1≈x as x increasesy=x
To find vertical asymptote equate denominator to zero.
x−1=0x=1
Hence, the oblique asymptote is y=x , vertical asymptote x=1
(d)
To plot: the given function.
(d)

Explanation of Solution
Given information:
Consider f(x)=x2−x+1x−1
Graph:
The graph of the f(x)=x2−x+1x−1 is shown below
Chapter 2 Solutions
EBK PRECALCULUS W/LIMITS
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