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.

To find the domain of a function, we need to consider any restrictions on the independent variable (in this case, x) that would cause the function to be undefined. In the given function, g(x) = 12 Ln x / x, we have two potential restrictions:

1. The natural logarithm function, Ln(x), is only defined for positive values of x. So, we need to ensure that x > 0.
2. Additionally, the denominator, x, cannot be equal to 0 since division by zero is undefined. Therefore, x โ‰  0.

Considering both restrictions, we can conclude that the domain of g(x) is (0, โˆž), meaning x can take any positive value but cannot be zero.

Now let's discuss how to determine the intervals of increase and decrease of the function and approximate any relative maximum or minimum values.

The first step is to find the derivative of the function. In this case, we can use the quotient rule to find the derivative of g(x). The derivative of Ln x is 1/x, and the derivative of x is 1. Applying the quotient rule, we have:

g'(x) = [(1/x) * x - Ln x * 1] / x^2
= (1 - Ln x) / x^2

Next, we need to find the critical points of the function, where g'(x) = 0 or is undefined. In this case, g'(x) is always defined, so we only need to find where g'(x) = 0, which corresponds to:

(1 - Ln x) / x^2 = 0

Simplifying, we get 1 - Ln x = 0, which means Ln x = 1. By taking the exponential of both sides, we find that x = e.

So, the critical point of the function is x = e.

To determine the intervals of increase and decrease, we can use the first derivative test. We choose test intervals less than e (e.g., x = 1) and greater than e (e.g., x = 3). By evaluating g'(x) in these intervals, we can determine if the derivative is positive (increasing) or negative (decreasing).

For x = 1, g'(1) = (1 - Ln 1) / 1^2 = 1, which is positive. So, the function is increasing for values less than e.

For x = 3, g'(3) = (1 - Ln 3) / 3^2, which is negative. So, the function is decreasing for values greater than e.

Therefore, the intervals of increase and decrease are as follows:

Increasing: (0, e)
Decreasing: (e, โˆž)

As for any relative maximum or minimum values, we can look for the presence of an inflection point, which would indicate a change in concavity. However, for this function, we only have one critical point (x = e), and no further information has been provided, such as the behavior of the function towards positive or negative infinity. Without this information or any other discontinuities, it is not possible to determine the presence of a relative maximum or minimum value.