What are all real values of a if f(x)=18x^2-lnx >= a, x>0 ?
We need to find the real values of a for which f(x)=18x^2 - lnx >= a
a is the minimum value of the function f(x) = 18x^2 - ln x.
At the minimum value of f(x) = 18x^2 - ln x, f'(x) = 0 and f''(x) is positive.
f'(x) = 36x - 1/x = 0
=> 36x^2 - 1 = 0
=> (6x - 1)(6x + 1) = 0
=> x = 1/6 and x = (-1/6)
f''(x) = 72x, this is positive for x = (1/6), so the minimum value lies at x = (1/6)
At x = 1/6, f(x) = 18x^2 - lnx = 18*(1/36) - ln (1/6)
=> 1/2 - ln (1/6)
So the values of a are a <= (1/2 - ln(1/6))
The required values that a can take are a < (1/2 - ln(1/6))
The first thing to do is to analyze the monotony of the function f(x). For this reason, we'll differentiate the function:
f'(x) = 36x - 1/x
We'll put f'(x) = 0.
36x - 1/x = 0
(36x^2 - 1)/x = 0
We notice that we have a difference of 2 squares at numerator:
(36x^2 - 1) = (6x - 1)(6x + 1)
f'(x) = 0 if and only if (6x - 1)(6x + 1) = 0.
6x - 1 = 0 => x = 1/6
6x + 1 = 0
x = -1/6
The derivative is negative, over the range (0;1/6) and it is positive over the range (1/6 ; +infinite).
That means that the function is decreasing over the interval (0;1/6] and it is increasing over the range [1/6 ; +infinite).
So, the point f(1/6) is a local minimum point for the function.
f(1/6) = 18/36 - ln(1/6) = 1/2 + ln 6
So, a =< 1/2 + ln 6
All real values of "a" are located in the interval (-infinite ; 1/2 + ln6].