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.
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
To determine the domain of a function, we need to find the values of x for which the function is defined. In this case, the function g(x) = (12 ln x) / x involves natural logarithm (ln) and a division by x.
The natural logarithm function is defined for positive values of x, so ln x is only defined for x > 0. Additionally, the denominator x cannot be equal to zero because division by zero is undefined. Therefore, the domain of the function is x > 0, or written interval notation, (0, ∞).
To determine the intervals in which the function is increasing or decreasing, we need to examine the derivative of the function.
Let's take the derivative of g(x) to find the critical points:
g'(x) = (12/x) - (12 ln x)/(x^2)
To find the critical points, we set the derivative equal to zero and solve for x:
(12/x) - (12 ln x)/(x^2) = 0
To simplify the equation, we will multiply through by x^2:
12x - 12 ln x = 0
Dividing through by 12:
x - ln x = 0
We can solve this equation graphically or numerically. Using a graphing utility or calculator, graph the equation y = x - ln x, and find the intersection point(s) with the x-axis.
For the given function g(x) = (12 ln x) / x, there is only one critical point, which occurs at x = 1. Therefore, there is no interval of increasing or decreasing.
To determine if there are any relative maximum or minimum values, we can examine the second derivative of the function.
g''(x) = (12 - 24 ln x) / (x^3)
To find the points where the concavity changes, we set the second derivative equal to zero and solve for x:
(12 - 24 ln x) / (x^3) = 0
Simplifying the equation:
12 - 24 ln x = 0
-24 ln x = -12
Dividing through by -24:
ln x = 1/2
x = e^(1/2)
Using a calculator, we find x ≈ 1.6487
Substituting this value into the original function:
g(1.6487) ≈ 9.162
So, there is a relative maximum at x ≈ 1.6487 with a value of approximately 9.162.
To summarize:
Domain: x > 0, or (0, ∞)
Increasing/Decreasing: The function does not have any intervals of increasing or decreasing.
Relative Maximum: (1.6487, 9.162)