Given the function `f(x)=1/(x+2)` . Find the points of discontinuity of the function `f(f(x))`

Explanation

To find the points of discontinuity of the function \( f(f(x)) \) where \( f(x) = \frac{1}{x + 2} \), we will follow these steps:

### Step 1: Identify the points of discontinuity of \( f(x) \)

The function \( f(x) \) is defined as:
\[ f(x) = \frac{1}{x + 2} \]
For \( f(x) \) to be discontinuous, the denominator must be zero:
\[ x + 2 = 0 \]
Solving for \( x \):
\[ x = -2 \]
Thus, \( x = -2 \) is a point of discontinuity for \( f(x) \).

### Step 2: Find \( f(f(x)) \)

Next, we need to find \( f(f(x)) \):
\[ f(f(x)) = f\left(\frac{1}{x + 2}\right) \]
Substituting \( \frac{1}{x + 2} \) into \( f(x) \):
\[ f(f(x)) = \frac{1}{\frac{1}{x + 2} + 2} \]
To simplify this, we need a common denominator for the terms in the denominator:
\[ f(f(x)) = \frac{1}{\frac{1 + 2(x + 2)}{x + 2}} = \frac{1}{\frac{1 + 2x + 4}{x + 2}} = \frac{x + 2}{2x + 5} \]

### Step 3: Identify the points of discontinuity of \( f(f(x)) \)

Now, we need to find the points of discontinuity of \( f(f(x)) \):
\[ f(f(x)) = \frac{x + 2}{2x + 5} \]
For \( f(f(x)) \) to be discontinuous, the denominator must be zero:
\[ 2x + 5 = 0 \]
Solving for \( x \):
\[ 2x = -5 \implies x = -\frac{5}{2} \]

### Step 4: Combine the points of discontinuity

The points of discontinuity for \( f(f(x)) \) are:
1. From \( f(x) \): \( x = -2 \)
2. From \( f(f(x)) \): \( x = -\frac{5}{2} \)

Thus, the points of discontinuity of the function \( f(f(x)) \) are:
\[ x = -2 \quad \text{and} \quad x = -\frac{5}{2} \]

### Summary of Points of Discontinuity
- \( x = -2 \)
- \( x = -\frac{5}{2} \)

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