Find the critical numbers of f(lf any). Find the open intervals on which the fun
ID: 2866868 • Letter: F
Question
Find the critical numbers of f(lf any). Find the open intervals on which the function is increasing or decreasing and locate all relative extrema. Use a graphing utility to confirm your results. f(x) = (x + 9)^2/3 Critical numbers: (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.) Increasing: (-infinity, -9) (-9, infinity) (-infinity, 0) (-10, infinity) none of these Decreasing: (-infinity, -9) ( -9,infinity) (-infinity, 0) (-10,infinity) none of these Relative extrema: (If an answer does not exist, enter DNE.) relative maximum (x, y) = relative minimum (x, y) =Explanation / Answer
f(x) = (x + 9)^(2/3)
To find criticals, we have to derive the function and equate derivative to 0
f'(x) = (2/3) * (x + 9)^(-1/3) = 0
(x + 9)^(-1/3) = 0
The above expression has an exponent of -1/3, a negative value
So, no critical numbers exist.
DNE ---> FIRST ANSWER
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Increasing/Decreasing :
f'(x) = (2/3) * (x + 9)^(-1/3)
From this, we notice that there is a vertical asymptote at x = -9
So, that splits the x intervals into (-inf , -9) and (-9 , inf)
Region 1 : (-inf , -9)
Testvalue = -10
f'(x) = (2/3) * (x + 9)^(-1/3)
f'(-10) = (2/3) * (-10 + 9)^(-1/3)
f'(-10) = (2/3) * (-1)
f'(-10) = -1 --> negative
So, decreasing over this region
Region 2 : (-9 , inf)
Testvalue = 0
f'(0) = (2/3) * (0 + 9)^(-1/3)
f'(0) = some positive value
So, increasing over this region
So, increasing : (-9 , inf) ----> SECOND ANSWER, option 2
Decreasing : (-inf , -9) ---> THIRD ANSWER, option 1
Relative extrema :
We have found that there are no critical numbers
So, there are no relative or local extrema
So, relative maximum = DNE ---> FOURTH ANSWER
relative minimum = DNE ---> FIFTH ANSWER
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