given the function f(x) = \\frac{x^{2}-27}{x - 6}, find the absolute minimum value on the interval 6, 12.

Explanation:

Step1: Find the derivative

Using the quotient - rule $\left(\frac{u}{v}\right)'=\frac{u'v - uv'}{v^{2}}$, where $u = x^{2}-27$, $u' = 2x$, and $v=x - 6$, $v'=1$. Then $f'(x)=\frac{2x(x - 6)-(x^{2}-27)\times1}{(x - 6)^{2}}=\frac{2x^{2}-12x-x^{2}+27}{(x - 6)^{2}}=\frac{x^{2}-12x + 27}{(x - 6)^{2}}$.

Step2: Find the critical points

Set $f'(x)=0$, so $x^{2}-12x + 27 = 0$. Factoring gives $(x - 3)(x - 9)=0$, and the solutions are $x = 3$ and $x = 9$. But $x = 3$ is not in the interval $[6,12]$, so we only consider $x = 9$.

Step3: Evaluate the function at critical and end - points

Evaluate $f(x)$ at $x = 6$, $x=9$, and $x = 12$. The function $f(x)=\frac{x^{2}-27}{x - 6}$ is undefined at $x = 6$ (vertical asymptote). $f(9)=\frac{9^{2}-27}{9 - 6}=\frac{81-27}{3}=\frac{54}{3}=18$. $f(12)=\frac{12^{2}-27}{12 - 6}=\frac{144-27}{6}=\frac{117}{6}=\frac{39}{2}=19.5$.

Answer:

$18$