\( 1 \leftarrow \quad \) Find the \( x \)-values of all points where the function has any relative extrema. Find the value(s) of any relative extrema. \[ f(x)=\frac{x^{2}-4 x+4}{x-6} \]

To find the \( x \)-values of all points where the function has any relative extrema, we need to find the critical points of the function. The function given is: \[ f(x) = \frac{x^{2}-4x+4}{x-6} \] To find the critical points, we need to find the values of \( x \) where the derivative of the function is equal to zero or undefined. Let's find the derivative of the function using the quotient rule: \[ f'(x) = \frac{(x-6)(2x-4) - (x^{2}-4x+4)(1)}{(x-6)^{2}} \] Now, we need to find the values of \( x \) where \( f'(x) = 0 \) or \( f'(x) \) is undefined. Let's simplify the derivative: \[ f'(x) = \frac{(x-6)(2x-4) - (x^{2}-4x+4)}{(x-6)^{2}} \] Now, we need to find the values of \( x \) where \( f'(x) = 0 \) or \( f'(x) \) is undefined. Let's solve for \( f'(x) = 0 \): \[ (x-6)(2x-4) - (x^{2}-4x+4) = 0 \] Solving this equation will give us the critical points of the function. Once we have the critical points, we can determine the relative extrema by evaluating the function at these points. Let's solve the equation \( (x-6)(2x-4) - (x^{2}-4x+4) = 0 \) to find the critical points. Solve the quadratic equation by following steps: - step0: Solve by factoring: \(\left(x-6\right)\left(2x-4\right)-\left(x^{2}-4x+4\right)=0\) - step1: Remove the parentheses: \(\left(x-6\right)\left(2x-4\right)-x^{2}+4x-4=0\) - step2: Expand the expression: \(x^{2}-12x+20=0\) - step3: Factor the expression: \(\left(x-10\right)\left(x-2\right)=0\) - step4: Separate into possible cases: \(\begin{align}&x-10=0\\&x-2=0\end{align}\) - step5: Solve the equation: \(\begin{align}&x=10\\&x=2\end{align}\) - step6: Rewrite: \(x_{1}=2,x_{2}=10\) The critical points of the function are \( x = 2 \) and \( x = 10 \). Now, we need to evaluate the function at these critical points to determine the relative extrema. Let's substitute \( x = 2 \) and \( x = 10 \) into the function \( f(x) = \frac{x^{2}-4x+4}{x-6} \) to find the relative extrema. Evaluate the expression by following steps: - step0: Evaluate: \(\frac{x^{2}-4x+4}{x-6}\) - step1: Substitute: \(\frac{2^{2}-4\times 2+4}{2-6}\) - step2: Multiply the numbers: \(\frac{2^{2}-8+4}{2-6}\) - step3: Calculate: \(\frac{0}{2-6}\) - step4: Subtract the numbers: \(\frac{0}{-4}\) - step5: Divide the terms: \(0\) Substitute \( x=10 \) into the expression \( \frac{x^{2}-4x+4}{x-6} \). Evaluate the expression by following steps: - step0: Evaluate: \(\frac{x^{2}-4x+4}{x-6}\) - step1: Substitute: \(\frac{10^{2}-4\times 10+4}{10-6}\) - step2: Multiply the numbers: \(\frac{10^{2}-40+4}{10-6}\) - step3: Calculate: \(\frac{64}{10-6}\) - step4: Subtract the numbers: \(\frac{64}{4}\) - step5: Divide the terms: \(16\) The relative extrema of the function are: - At \( x = 2 \), the function has a relative minimum of 0. - At \( x = 10 \), the function has a relative maximum of 16.