f(x) = 18x^2 - lnx

First let us find f'(x):

f'(x) = 36x - (1/x)

= (36x^2 - 1)/x

= (6x-1)(6x+1)/x

Now let us determine the critical values:

(6x-1)(6x+1)/x = 0

==> x1= 1/6

==> x2= -1/6

Also , the function is not defined for x =0

==> we have the intervals:

(-inf,-1/6), (-1/6,0) (0,1/6)(1/6, inf)

For (-inf, -1/6) ==> f'(x) < 0 ==> f is decreasing

For (-1/6, 0) ==> f'(x) > 0 ==> f is increasing

for (0, 1/6) ==> f'(x) < 0 ==> f is decreasing

for (1/6, inf) ==> f'(x) > 0==> f is increasing

f(x) = 18x^2-lnx

f'(x) = 36x-1/x.

When f'(x) > 0 in (36x-1/x) > 0. O 36 x^2-1) > 0. Or

(6x-1)(6x+1) > 0

Or x does not belong to [-1/6 , 1/6] , f'(x) > 0 and f(x) is monotonously increasing .

When x belogs to { -1/6 , 1/6), f'(x) < 0 and f(x) is monotonously decreasing.

To determine the inervals of monotony of a function, we have to calculate it's derivative and to determine the intervals where the derivative is positive or negative.

If derivative is positive over an interval, then the function is increasing over that interval.

If derivative is negative over an interval, then the function is decreasing over that interval.

Now, we'll calculate the derivative of the function:

f'(x) = (18x^2 - ln x)'

f'(x) = 18*2*x - 1/x

f'(x) = 36x - 1/x

We'll calculate the LCD:

f'(x) = (36x^2-1)/x

We'll verify where the numerator is positive or negative, if x>0 (the constraint of the existance of logarithm).

36x^2-1 = 0

It is a difference of squares:

(6x-1)(6x+1) = 0

We'll put each factor from the product as being 0.

6x-1 = 0

We'll add 1 both sides:

6x = 1

We'll divide by 6;

x = 1/6

6x+1 = 0

6x = -1

x = -1/6

Now, we'll discuss the values of f'(x), but only for x values bigger than 1/6, because -1/6 does not belong to the admissible interval (0,+inf.).

f'(x) is positive over the interval [1/6, +inf.) and is negative over the interval (0,1/6).

The function is increasing over [1/6, +inf.) and is decreasing (0,1/6).