This is the graph of function y = 1 / (3x - 2) Indicate the transition points an
ID: 2867397 • Letter: T
Question
This is the graph of function y = 1 / (3x - 2)
Indicate the transition points and asymptotes. (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.)
(a) What is the local minima?
(b) What is the local maxima?
(c) What are the inflection points?
(d) What are the vertical asymptotes?
(e) What are the horizontal asymptotes?
This is the graph of function y = 1 / (3x - 2) Indicate the transition points and asymptotes. (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.) (a) What is the local minima? (b) What is the local maxima? (c) What are the inflection points? (d) What are the vertical asymptotes? (e) What are the horizontal asymptotes?Explanation / Answer
y = 1 / (3x - 2)
Deriving :
y' = -3 / (3x - 2)^2
First we find critical points, so y' = 0
-3 / (3x - 2)^2 = 0
-3 = 0 ---> ABSURD
So, no critical points and therefore, no extrema
Local minima : DNE
Local maxima : DNE
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Inflection :
y' = -3 / (3x - 2)^2
Deriving again :
y'' = 18 / (3x - 2)^3
To find inflection, y'' = 0
18 / (3x - 2)^3 = 0
Crossmultiplying :
18 = 0 ---> ABSURD
So, no inflection points exist
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Vertical asymptotes :
Equate denominator to 0
3x - 2 = 0
3x = 2
x = 2/3 ---> VA
Horizontal asymptote :
Degree of numerator = ZERO
Degree of denominator = ONE
Deg of numerator LESS THAN deg of denominator
So, the horizontal asymptote is : y = 0
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Here are the answers :
Local minima : DNE
Local maxima : DNE
Inflection points : DNE
Vertical asymptotes : x = 2/3
Horizontal asymptote : y = 0
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