Consider the function [f(x) = { 2 x + 7 }/ { 6 x + 3 }] . For this function ther

Question

Consider the function [f(x) = { 2 x + 7 }/ { 6 x + 3 }] . For this function there are two important intervals: [(-infty, A)] and [(A,infty)] where the function is not defined at [A] .
Find [A]   
For each of the following intervals, tell whether [f(x)] is increasing (type in INC) or decreasing (type in DEC).
[(-infty, A)] :  
[(A,infty)]   
Note that this function has no inflection points, but we can still consider its concavity. For each of the following intervals, tell whether [f(x)] is concave up (type in CU) or concave down (type in CD).
[(-infty, A)] :  
[(A,infty)]

Explanation / Answer

f(x) = (2x + 7) / (6x + 3)

Equate denominator to 0

6x + 3 = 0

6x = -3

x = -3/6

x = -1/2

So, A = -1/2 ----> ANSWER

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Increase/Decrease :

f(x) = (2x + 7) / (6x + 3)

Deriving :

u = 2x + 7 , v = 6x + 3
u' = 2 , v' =6

(u'v - uv') / v^2, by quotient rule

(12x + 6 - 12x - 42) / (6x + 3)^2

f'(x) = -36 / (6x + 3)^2

But we have a VA at x = -1/2

So, this splits the region into (-inf , -1/2) and (-1/2 , inf).....

Region 1 : (-inf , -1/2)
Testvalue = -1
f'(x) = -36 / (6x + 3)^2
f'(-1) = negative
So, DECREASING over (-inf , -1/2)

Region 2 : (-1/2 , inf)
Testvalue = 0
f'(x) = -36 / (6x + 3)^2
f'(0) = -36 / (0 + 3)^2 --> -4 --> negative
So, DECREASING over (-1/2 , inf)

So, decreasing over : (-inf , -1/2) U (-1/2 , inf)
Increasing : nowhere

(-inf , A) : decreasing ---> ANSWER
(A , inf) : decreasing ---> ANSWER

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Concavity :

f'(x) = -36 / (6x + 3)^2

f''(x) = -36 * -2 / (6x + 3)^3 * 6

f''(x) = 432 / (6x + 3)^3

But undefined at x = -1/2

Region 1 : (-inf , -1/2)
Testvalue = -1
f''(x) = 432 / (6x + 3)^3
f''(-1) = 432 / (-3)^3 --> negative
So, concave down

Region 2 : (-1/2 , inf)
Testvalue = 0
f''(0) = 432 / (0 + 3)^3 --> positive
So, concave up

So, answers :

(-inf , A) : concave down ---> ANSWER
(A , inf) : concave up ---> ANSWER

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