If the function `f(x) = 1/(x+2)`, then find the points of discountinuity of the composite function ` y = f{f(x)}`.

Explanation

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

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

The function \( f(x) = \frac{1}{x+2} \) is discontinuous when its denominator is zero. Therefore, we set the denominator equal to zero:

\[ x + 2 = 0 \]

Solving for \( x \):

\[ x = -2 \]

Thus, \( f(x) \) is discontinuous at \( x = -2 \).

### Step 2: Find the composite function \( y = f(f(x)) \)

Next, we need to find the composite function \( y = f(f(x)) \). We first substitute \( f(x) \) into itself:

\[ f(f(x)) = f\left(\frac{1}{x+2}\right) \]

Now, we substitute \( \frac{1}{x+2} \) into the function \( f \):

\[ f\left(\frac{1}{x+2}\right) = \frac{1}{\frac{1}{x+2} + 2} \]

### Step 3: Simplify the expression

To simplify \( \frac{1}{\frac{1}{x+2} + 2} \), we first find a common denominator for the terms in the denominator:

\[ \frac{1}{x+2} + 2 = \frac{1}{x+2} + \frac{2(x+2)}{x+2} = \frac{1 + 2(x+2)}{x+2} = \frac{1 + 2x + 4}{x+2} = \frac{2x + 5}{x+2} \]

Thus, we have:

\[ f(f(x)) = \frac{1}{\frac{2x + 5}{x + 2}} = \frac{x + 2}{2x + 5} \]

### Step 4: Identify points of discontinuity for \( y = f(f(x)) \)

The function \( y = \frac{x + 2}{2x + 5} \) is discontinuous when its denominator is zero. Therefore, we set the denominator equal to zero:

\[ 2x + 5 = 0 \]

Solving for \( x \):

\[ 2x = -5 \implies x = -\frac{5}{2} \]

### Step 5: List all points of discontinuity

The points of discontinuity for the composite function \( y = f(f(x)) \) are:

1. From the original function \( f(x) \): \( x = -2 \)
2. From the composite function \( f(f(x)) \): \( x = -\frac{5}{2} \)

Thus, the points of discontinuity are \( x = -2 \) and \( x = -\frac{5}{2} \).

### Final Answer

The points of discontinuity of the composite function \( y = f(f(x)) \) are \( x = -2 \) and \( x = -\frac{5}{2} \).

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