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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.
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.
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