Locate any relative extrema and inflection points.
y = (ln x)/x
To locate the relative extrema and inflection points of the function y = (ln x)/x, we need to find the critical points and analyze the behavior of the function around those points.
Step 1: Find the derivative of the function.
To find the critical points, we first need to find the derivative of the function. The derivative of y = (ln x)/x can be found using the quotient rule:
dy/dx = (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: Find the critical points.
To find the critical points, we set the derivative equal to zero and solve for x:
(1 - ln x) / x^2 = 0
To determine when the numerator is equal to zero, we solve the equation 1 - ln x = 0:
ln x = 1
x = e
Now, we check for any potential asymptotes by checking if the denominator is equal to zero:
x^2 ≠ 0 for all x
Therefore, the critical point is x = e.
Step 3: Analyze the behavior around the critical point.
To analyze the behavior around the critical point x = e, we can use the second derivative test. Let's find the second derivative:
d²y/dx² = d/dx[(1 - ln x) / x^2]
= (d/dx(1 - ln x) * x^2 - (1 - ln x) * d/dx(x^2)) / x^4
= (0 - (1 - ln x) * 2x) / x^4
= (ln x - 1) * 2x / x^4
= (2x * ln x - 2x) / x^4
= 2ln x / x^3 - 2 / x^3
To determine the nature of the critical point, evaluate the second derivative at x = e:
d²y/dx² = 2ln e / e^3 - 2 / e^3
= 2(1) / e^3 - 2 / e^3
= 0
The second derivative at x = e is zero.
Step 4: Determine the extrema and inflection points.
Based on the second derivative test, if the second derivative is:
- Positive, then we have a relative minimum.
- Negative, then we have a relative maximum.
- Zero, then the test is inconclusive.
Since the second derivative at x = e is zero, the second derivative test is inconclusive. Therefore, we need to further analyze the behavior of the function.
To investigate the behavior of the function for x < e and x > e separately, we can check the signs of the first derivative:
For x < e:
Since dy/dx = (1 - ln x) / x^2, the sign of dy/dx is positive for x < e. This means that the function is increasing for x < e.
For x > e:
Since dy/dx = (1 - ln x) / x^2, the sign of dy/dx is negative for x > e. This means that the function is decreasing for x > e.
Inflection Points:
To find the inflection points, we need to check where the concavity of the function changes. We can do this by checking the sign changes of the second derivative.
For the second derivative, we have:
d²y/dx² = 2ln x / x^3 - 2 / x^3
To find where the second derivative changes sign, we need to solve the equation d²y/dx² = 0:
2ln x / x^3 - 2 / x^3 = 0
Simplifying the equation:
2ln x - 2 = 0
ln x = 1
x = e
Therefore, the inflection point is x = e.
In summary, for the function y = (ln x)/x:
- There is a relative extremum at x = e (the critical point).
- There is an inflection point at x = e (the critical point).
- The function is increasing for x < e.
- The function is decreasing for x > e.