Find the relative extrema of f(x)= x/ln(x)
To find the relative extrema of a function, you need to determine where the function's derivative equals zero or is undefined. We'll start by finding the derivative of f(x) using the quotient rule.
Given the function f(x) = x/ln(x), let's find its derivative:
f'(x) = [x * (d/dx)(ln(x)) - ln(x) * (d/dx)(x)] / (ln(x))^2
To evaluate (d/dx)(ln(x)), we use the chain rule. The derivative of ln(x) is 1/x, and the derivative of x with respect to x is 1. So,
(d/dx)(ln(x)) = 1/x
Substituting this value back into f'(x):
f'(x) = [x * (1/x) - ln(x) * 1] / (ln(x))^2
= [1 - ln(x)] / (ln(x))^2
Now, to find the points where f'(x) = 0 or is undefined, we'll set the numerator equal to zero and solve for x:
1 - ln(x) = 0
ln(x) = 1
Using the property of logarithms, rewrite the equation as:
x = e
So, the only critical point where f'(x) = 0 is x = e.
Next, we should check the endpoints of any closed intervals where the function is defined. In this case, there are no endpoints because x ranges from 0 to infinity.
Now, we need to determine the nature of the critical point x = e by examining whether f'(x) changes sign at that point. To do this, we evaluate f'(x) for a value in each interval on both sides of x = e.
When x < e:
Let's choose x = 1 (a value less than e)
f'(1) = [1 - ln(1)] / (ln(1))^2 = 1
Since f'(1) = 1 > 0, we know that f(x) is increasing on the interval (-∞, e).
When x > e:
Let's choose x = 3 (a value greater than e)
f'(3) = [1 - ln(3)] / (ln(3))^2 ≈ -0.184
Since f'(3) ≈ -0.184 < 0, we know that f(x) is decreasing on the interval (e, ∞).
Based on the information gathered, we can conclude that x = e is a relative minimum of f(x) = x/ln(x). There are no other relative extrema.