Determine the intervals where the function is increasing and where it is decreasing.

f(x)=(ln(x))/x

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Determine the intervals where the function is increasing and where it is decreasing.

f(x)=(ln(x))/x

Look for places where the derivative f'(x) is positive or negative.

f'(x) = (1 - lnx)/x^2

f'(x) is positive where lnx < 1

lnx = 1 at x = e

at x>e, the derivative is negative and the function decreases.

The function f(x) is not defined for x = 0 or x<0

Check:

at x = 0.1, f(x) = -2.303
at x = 0.5, f(x) = -1.386
at x = 1, f(x) = 0
at x = 2, f(x) = 0.3466
at x = e = 2.71828, f(x) = 0.3686
at x = 3, f(x) = 0.3662

To determine the intervals where the function f(x) = (ln(x))/x is increasing or decreasing, we need to find the first derivative of the function and then analyze its sign.

Step 1: Find the first derivative, f'(x), using the quotient rule of differentiation.

f'(x) = [(x * d/dx(ln(x))) - (ln(x) * d/dx(x))] / (x^2)
= [(x * (1/x)) - (ln(x) * 1)] / (x^2)
= (1 - ln(x))/x^2

Step 2: Analyze the sign of f'(x) to determine the intervals of increase and decrease.

To do this, we need to find the critical points by setting f'(x) = 0 and solve for x.

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

This equation is satisfied when 1 - ln(x) = 0, which means ln(x) = 1. Solving for x gives:

x = e, where e is the base of the natural logarithm.

So, the critical point is x = e.

Step 3: Test the intervals on either side of the critical point.

To test the intervals, choose a value smaller than e, a value between 1 and e, and a value larger than e.

For x < e:
Choose x = 1. Plugging this into f'(x), we get:
f'(1) = (1 - ln(1))/1^2 = (1 - 0)/1 = 1.
Since f'(1) > 0, this means that the function f(x) = (ln(x))/x is increasing for x < e.

For x > e:
Choose x = 10. Plugging this into f'(x), we get:
f'(10) = (1 - ln(10))/10^2 ≈ (1 - 2.30259)/100 ≈ -0.01303.
Since f'(10) < 0, this means that the function f(x) = (ln(x))/x is decreasing for x > e.

Therefore, the function f(x) = (ln(x))/x is increasing for x < e (where x is less than the base of the natural logarithm, e), and it is decreasing for x > e (where x is greater than the base of the natural logarithm, e).