We have:
[MATH]\frac{d}{dx}\left(x^4-2x^3-3x+9\right)=4x^3-6x^2-3[/MATH]
There is only 1 sign change, so this means there will be one real root. To determine this one real root, I would use a numeric root finding technique to get an approximation. Are you familiar with the Newton-Raphson method? Depending on your professor, you may simply be allowed to use a CAS (computer algebra system) like WolframAlpha to approximate the root. What do you find?
Using W|A, I find the one real root is approximately:
[MATH]x\approx1.74601665058591[/MATH]
Now, if we define:
[MATH]f(x)=x^4-2x^3-3x+9[/MATH]
We find:
[MATH]f(1.74601665058591)=2.41003192808928105>0[/MATH]
Since [MATH]\lim_{x\to\pm\infty}f(x)=+\infty[/MATH]
We may conclude:
[MATH]f_{\min}\approx2.41003192808928105[/MATH]
Thus \(f(x)\) has no real roots, and is positive over its entire domain. And so we revisit:
[MATH]\frac{x^4-2x^3-3x+9}{3(x-2)}>0[/MATH]
The expression on the LHS will thus have the same sign as:
[MATH]x-2[/MATH]
which means the solution for the inequality is:
[MATH]x-2>0[/MATH]
or:
[MATH]x>2[/MATH]
as you found. But you stated you were trying to find the extrema of:
[MATH]g(x)=\ln|x-2|+\frac{1}{12}x^4-x+3[/MATH]
And what we want is critical values, that is values of \(x\) such that \(g'(x)=0\), we as we now know, there are none. This means there are no local extrema, and we must examine the behavior of the function at the boundaries of its domain. Can you give the domain?