If `f(x)=(x+2)/(x-2),AA x!=2`, value of `f'(-2)=`

Options

A. 1
B. `1/4`
C. 0
D. `-1/4`

Answer

To find the value of \( f'(-2) \) for the function \( f(x) = \frac{x+2}{x-2} \), we will follow these steps:

### Step 1: Differentiate the function using the Quotient Rule
The Quotient Rule states that if you have a function \( f(x) = \frac{g(x)}{h(x)} \), then the derivative \( f'(x) \) is given by:
\[ f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{[h(x)]^2} \]
In our case, let:
- \( g(x) = x + 2 \) (numerator)
- \( h(x) = x - 2 \) (denominator)

### Step 2: Calculate \( g'(x) \) and \( h'(x) \)
- The derivative \( g'(x) = 1 \) (since the derivative of \( x \) is 1 and the derivative of a constant is 0).
- The derivative \( h'(x) = 1 \) (similarly).

### Step 3: Apply the Quotient Rule
Now we can apply the Quotient Rule:
\[ f'(x) = \frac{(1)(x - 2) - (x + 2)(1)}{(x - 2)^2} \]
This simplifies to:
\[ f'(x) = \frac{x - 2 - (x + 2)}{(x - 2)^2} \]

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

### Step 5: Substitute \( x = -2 \) into the derivative
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:
\[ \boxed{-\frac{1}{4}} \]