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What is the domain? Use a graphing utility to determine the intervals in which the function is increasing and decreasing and ap...Asked by Mark
What is the domain? Use a graphing utility to determine the intervals in which the function is increasing and decreasing and approximate any relative maximum or minimun values of the function.
g(x) = 12 Ln x / x
My answer was: Domain = (12,0)
decreasing = (3,-2) and increasing = (-2,3) but these were wrong.
Answers
Answered by
Reiny
First of all we can only take the ln of positive numbers,
so x > 0
(notice that also takes care of the possibility of dividing by zero)
g'(x) using the quotient rule was
(12 - 12lnx)/x^2
let's set that equal to zero for max/mins
12lnx = 12
lnx = 1
x = e
So when x = e the curve is neither increasing nor decreasing.
We could take the second derivative , but an easier way is to simply test a value less than and a value greater than e, and see if it is positive or negative.
We don't need that actual value
let x = 2
g'(2) = (12 - 12ln2)/4 , clearly positive , since ln2 is negative)
let x = 3
g'(3) = (12-l2ln3)/9 , which is negative
also if x = 3
g(e) = 12lne/e = 12/e = appr. 12/3 = appr.4
(calculator value =4.4145)
so...
domain : x > 0
increasing : 0 < x < e
decreasing: x > e
so x > 0
(notice that also takes care of the possibility of dividing by zero)
g'(x) using the quotient rule was
(12 - 12lnx)/x^2
let's set that equal to zero for max/mins
12lnx = 12
lnx = 1
x = e
So when x = e the curve is neither increasing nor decreasing.
We could take the second derivative , but an easier way is to simply test a value less than and a value greater than e, and see if it is positive or negative.
We don't need that actual value
let x = 2
g'(2) = (12 - 12ln2)/4 , clearly positive , since ln2 is negative)
let x = 3
g'(3) = (12-l2ln3)/9 , which is negative
also if x = 3
g(e) = 12lne/e = 12/e = appr. 12/3 = appr.4
(calculator value =4.4145)
so...
domain : x > 0
increasing : 0 < x < e
decreasing: x > e
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