Find x-value of all points where function has relative extrema and find the Values of these extrema.
f(x)= (x^2)/(ln(x))
f'(x) = x(2logx - 1)/(logx)^2
So, since f and f' are not defined at x=0,
f'(x) = 0 when
2logx = 1
logx = 1/2
x = √e
f(√e) = e/(1/2) = 2e
See the graph at
http://www.wolframalpha.com/input/?i=%28x^2%29%2F%28ln%28x%29%29+for+x+%3D+0..3
To find the x-values of all points where the function has relative extrema, we need to find the critical points first. Critical points occur where the derivative of the function is either zero or undefined.
Let's start by finding the derivative of the given function, f(x) = (x^2) / (ln(x)).
Using the quotient rule, the derivative of the function is given by:
f'(x) = [(2x * ln(x)) - (x^2 * (1/x))] / (ln(x))^2
Simplifying this expression further, we have:
f'(x) = (2x * ln(x) - x) / (ln(x))^2
Now, set f'(x) equal to zero to find the potential critical point(s):
(2x * ln(x) - x) / (ln(x))^2 = 0
Multiplying through by (ln(x))^2 to get rid of the denominator, we have:
2x * ln(x) - x = 0
Now, let's solve this equation for x:
x * (2ln(x) - 1) = 0
Either x = 0 or 2ln(x) - 1 = 0.
If x = 0, it is not valid since the logarithm of 0 is undefined.
Solving 2ln(x) - 1 = 0 for x:
2ln(x) = 1
ln(x) = 1/2
Using the property of logarithms, we can rewrite this equation as:
x = e^(1/2)
Hence, the critical point of the function is x = e^(1/2).
To determine the type of extrema at this critical point, we need to analyze the behavior of the function before and after the critical point.
Let's evaluate the function f(x) at x = e^(1/2) and check whether it has a relative maximum or minimum:
f(x) = (x^2) / (ln(x))
= (e^(1/2))^2 / ln(e^(1/2))
Simplifying further, we have:
f(x) = e / (ln(e^(1/2)))
= e / (1/2)
= 2e
Hence, at x = e^(1/2), the function has a relative maximum with the value of 2e.
Therefore, the function has one critical point at x = e^(1/2), which corresponds to a relative maximum with the value of 2e.