Given the function f(x)=18x^2-ln x, determine the number of real roots of the equation f(x)=m, if m is a real number.

First, we need to determine the local extremes of the
function. For this reason, we'll determine first the critical
points.

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

To
determine the critical points, we'll put f'(x)=0

36x - 1/x
= 0

36x^2 - 1 = 0

Since it is
a difference of 2 squares, we'll substitute it by the equivalent
product.

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

Now,
we'll set each factor as zero:

6x-1 =
0

x = 1/6

6x+1 =
0

x = -1/6

Since the function
f(x) does have a term ln x, we'll impose the constraint that the domain of the function
is (0 , +infinite).

Since the domain is (0,+infinite),
we'll reject the value x = -1/6.

The only critical point is
x = 1/6.

We'll calculate the 2nd derivative to see if the
extreme is a minimum or maximum point.

f"(x) = 36 +
1/x^2

It is obvious that f"(x)>0, so f(1/6) = 1/2 +
ln 6 is a minimum point.

Now, we'll discuss 3
cases:

1) m<1/2 + ln 6, the equation f(x) = m has no
real roots.

2)m = 1/2 + ln 6, then x = 1/6 is the only root
of the equation f(x) = m, since the function has just one minimum
point.

3) m>1/2 + ln 6, the equation f(x) = m has 2
real roots: x1 belongs to (0 ; 1/6) and x2 belongs to (1/6 ;
+infinite).

Conclusion: The equation could
have no roots for m<1/2 + ln 6, it could have one real root, for m = 1/2 + ln 6,
and it could have 2 real roots, located in the interval (0 ; 1/6) and [1/6 ; +infinte),
for m>1/2 + ln 6.