Question

question find the critical points of the function (f(x)=\frac{x - 6}{x^{2}-6x + 1}). if there is more than one critical point, write each value of (x) separated by a comma and enter an exact answer. provide your answer below.

Explanation:

Step1: Recall critical - point definition

Critical points occur where $f^{\prime}(x) = 0$ or $f^{\prime}(x)$ is undefined. First, use the quotient rule. If $f(x)=\frac{u(x)}{v(x)}$, then $f^{\prime}(x)=\frac{u^{\prime}(x)v(x)-u(x)v^{\prime}(x)}{v(x)^2}$. Here, $u(x)=x - 6$ and $v(x)=x^{2}-6x + 1$.
$u^{\prime}(x)=1$ and $v^{\prime}(x)=2x - 6$.
So, $f^{\prime}(x)=\frac{1\cdot(x^{2}-6x + 1)-(x - 6)\cdot(2x - 6)}{(x^{2}-6x + 1)^{2}}$.

Step2: Expand the numerator

Expand the numerator:
\[

$$\begin{align*} &(x^{2}-6x + 1)-(x - 6)(2x - 6)\\ =&x^{2}-6x + 1-(2x^{2}-6x-12x + 36)\\ =&x^{2}-6x + 1-(2x^{2}-18x + 36)\\ =&x^{2}-6x + 1-2x^{2}+18x - 36\\ =&-x^{2}+12x - 35 \end{align*}$$

\]
So, $f^{\prime}(x)=\frac{-x^{2}+12x - 35}{(x^{2}-6x + 1)^{2}}$.

Step3: Find where $f^{\prime}(x) = 0$

Set the numerator equal to 0: $-x^{2}+12x - 35 = 0$. Multiply through by - 1 to get $x^{2}-12x + 35 = 0$.
Factor the quadratic equation: $x^{2}-12x + 35=(x - 5)(x - 7)=0$.
Solving $(x - 5)(x - 7)=0$ gives $x = 5$ and $x = 7$.

Step4: Find where $f^{\prime}(x)$ is undefined

Set the denominator equal to 0: $x^{2}-6x + 1=0$. Using the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ for the quadratic equation $ax^{2}+bx + c = 0$. Here, $a = 1$, $b=-6$, $c = 1$.
$x=\frac{6\pm\sqrt{(-6)^{2}-4\times1\times1}}{2\times1}=\frac{6\pm\sqrt{36 - 4}}{2}=\frac{6\pm\sqrt{32}}{2}=\frac{6\pm4\sqrt{2}}{2}=3\pm2\sqrt{2}$.

Answer:

$3 - 2\sqrt{2},3 + 2\sqrt{2},5,7$