Final Answer:
![\(A = 1\) and \(B = e^{(1)/(6)}\). On the interval \((A, B]\), \(f(x)\) is increasing.](https://img.qammunity.org/2024/formulas/mathematics/high-school/zizq9ns0hk6tk1giewmcdfl1cyx7cx4njl.png)
Step-by-step explanation:
The critical points of the function 
 occur where the numerator or denominator equals zero or is undefined. In this case, the function is undefined at
occur where the numerator or denominator equals zero or is undefined. In this case, the function is undefined at 
 due to the natural logarithm, and the denominator becomes zero at
 due to the natural logarithm, and the denominator becomes zero at
 is a critical point.
 is a critical point.
To find B, set the denominator equal to zero and solve for x:
![\[x^6 = e^{(1)/(6)} \implies x = e^{(1)/(6)}.\]](https://img.qammunity.org/2024/formulas/mathematics/high-school/fxt6002oe64ubt9r06kh4a8ivn6nssgkbh.png)
So, 
 on the interval
 on the interval 
![\((A, B]\), we can analyze the sign of the derivative. Calculate \(f'(x)\)](https://img.qammunity.org/2024/formulas/mathematics/high-school/7qgppt3xj6srfq3uyes990dg49ppwzvp2l.png) using the quotient rule:
 using the quotient rule:
![\[f'(x) = (1 - 6\ln x)/(x^7).\]](https://img.qammunity.org/2024/formulas/mathematics/high-school/miivp5a8bsip1ce2dzmgu8l6n2mqoh5ex3.png)
On the interval 
![\((A, B]\), \(f'(x)\) is positive because \(\ln x < 1\) for \(x > 1\).](https://img.qammunity.org/2024/formulas/mathematics/high-school/mbszd82omtihtg7mj7c2vgoh4jivjvvcpz.png) Thus,
Thus, 
![\(f(x)\) is increasing on \((A, B]\).](https://img.qammunity.org/2024/formulas/mathematics/high-school/wvnwggyafp0068yy80jgixn5q1s1bxoktn.png)