Questions: What is the relative maximum and minimum for (x-1)/(x-2)^2 ?

Transcript text: What is the relative maximum and minimum for $\frac{x-1}{(x-2)^{2}}$ ?

Solution

Solution Steps

To find the relative maximum and minimum of the function $ f(x) = \frac{x-1}{(x-2)^2} $, we need to follow these steps:

Find the first derivative of the function, $ f'(x) $.
Determine the critical points by setting $ f'(x) = 0 $ and solving for $ x $.
Use the second derivative test to classify each critical point as a relative maximum, minimum, or neither.

Step 1: Find the First Derivative

We start with the function

\[ f(x) = \frac{x - 1}{(x - 2)^2} \]

To find the critical points, we first compute the first derivative $ f'(x) $:

\[ f'(x) = (x - 2)^{-2} - 2 \cdot \frac{x - 1}{(x - 2)^3} \]

Step 2: Determine Critical Points

Next, we set the first derivative equal to zero to find the critical points:

\[ f'(x) = 0 \]

Solving this equation, we find that the critical point is

\[ x = 0 \]

Step 3: Classify the Critical Point

To classify the critical point, we compute the second derivative $ f''(x) $:

\[ f''(x) = -\frac{4}{(x - 2)^3} + 6 \cdot \frac{x - 1}{(x - 2)^4} \]

Evaluating the second derivative at the critical point $ x = 0 $:

\[ f''(0) = -\frac{4}{(0 - 2)^3} + 6 \cdot \frac{0 - 1}{(0 - 2)^4} = -\frac{4}{-8} + 6 \cdot \frac{-1}{16} = \frac{1}{2} - \frac{3}{8} = \frac{4}{8} - \frac{3}{8} = \frac{1}{8} \]

Since $ f''(0) > 0 $, we conclude that $ x = 0 $ is a relative minimum.