If `f(x) = (x+2)/( x-2)` for all `x ne 2`, find `f' (-2)`.

Answer

To find \( f'(-2) \) for the function \( f(x) = \frac{x+2}{x-2} \), we will use the quotient rule for differentiation. The quotient rule states that if you have a function \( f(x) = \frac{u}{v} \), then the derivative \( f'(x) \) is given by:

\[ f'(x) = \frac{u'v - uv'}{v^2} \]

where \( u = x + 2 \) and \( v = x - 2 \).

### Step 1: Identify \( u \) and \( v \)
Let:
- \( u = x + 2 \)
- \( v = x - 2 \)

### Step 2: Differentiate \( u \) and \( v \)
Now we differentiate \( u \) and \( v \):
- \( u' = \frac{d}{dx}(x + 2) = 1 \)
- \( v' = \frac{d}{dx}(x - 2) = 1 \)

### Step 3: Apply the Quotient Rule
Using the quotient rule:
\[ f'(x) = \frac{u'v - uv'}{v^2} = \frac{(1)(x - 2) - (x + 2)(1)}{(x - 2)^2} \]

### Step 4: Simplify the Expression
Now simplify the numerator:
\[ f'(x) = \frac{x - 2 - (x + 2)}{(x - 2)^2} = \frac{x - 2 - x - 2}{(x - 2)^2} = \frac{-4}{(x - 2)^2} \]

### Step 5: Substitute \( x = -2 \)
Now we need to find \( f'(-2) \):
\[ f'(-2) = \frac{-4}{(-2 - 2)^2} = \frac{-4}{(-4)^2} = \frac{-4}{16} = -\frac{1}{4} \]

### Final Answer
Thus, the value of \( f'(-2) \) is:
\[ \boxed{-\frac{1}{4}} \]