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

Answer to Problem 51E
The domain of the given function is (−∞,0)∪(0,∞) .
Explanation of Solution
Given information:
Consider f(x)=2x2+1x
Calculation:
The domain is the set of x values for which the function is defined.
Find the undefined values of f(x)=2x2+1x
x=0
The function is define for all real values of x except at x=0
Hence, the domain is (−∞,0)∪(0,∞)
(b)
To identify: all the intercepts of the given function.
(b)

Answer to Problem 51E
No intercepts.
Explanation of Solution
Given information:
Consider f(x)=2x2+1x
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=f(x)=2x2+1x=0x is not real
x=0⇒y=f(0)=2(0)+1(0) undefined
Hence, no intercepts.
(c)
To find: the vertical or slant asymptotes of the given function.
(c)

Answer to Problem 51E
slant asymptote is y=2x ,vertical asymptote x=0
Explanation of Solution
Given information:
Consider f(x)=2x2+1x
Calculation:
Degree of numerator is greater than the degree of denominator. So there is no horizontal
Asymptote. Slant asymptote is there.
y=2x2+1x=2x+1x≈x as x increasesy=2x
To find vertical asymptote equate denominator to zero.
x=0
Hence, the oblique asymptote is y=2x ,vertical asymptote x=0
(d)
To plot: the given function.
(d)

Explanation of Solution
Given information:
Consider f(x)=2x2+1x
Graph:
The graph of the f(x)=2x2+1x is shown below
Chapter 2 Solutions
Precalculus with Limits
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