If `f (x) = x^(2) + 1,` then the value of (fof) (x) is equal to

Options

A. `x ^(4) + 1`
B. ` x ^(4) + 2x ^(2) + 2`
C. `x ^(4) + x ^(2) +1`
D. none of these

Answer

To find the value of \( (f \circ f)(x) \) where \( f(x) = x^2 + 1 \), we need to follow these steps:

### Step 1: Understand the notation \( (f \circ f)(x) \)
The notation \( (f \circ f)(x) \) means we need to apply the function \( f \) to itself. In other words, we will first find \( f(x) \) and then substitute that result back into \( f \).

### Step 2: Calculate \( f(x) \)
Given:
\[ f(x) = x^2 + 1 \]

### Step 3: Substitute \( f(x) \) into itself
Now we need to find \( f(f(x)) \):
\[ f(f(x)) = f(x^2 + 1) \]

### Step 4: Replace \( x \) in \( f(x) \) with \( x^2 + 1 \)
Now we substitute \( x^2 + 1 \) into the function \( f \):
\[ f(x^2 + 1) = (x^2 + 1)^2 + 1 \]

### Step 5: Expand \( (x^2 + 1)^2 \)
Now we need to expand \( (x^2 + 1)^2 \):
\[ (x^2 + 1)^2 = x^4 + 2x^2 + 1 \]

### Step 6: Add 1 to the expanded result
Now we add 1 to the result from the previous step:
\[ f(f(x)) = x^4 + 2x^2 + 1 + 1 = x^4 + 2x^2 + 2 \]

### Final Result
Thus, the value of \( (f \circ f)(x) \) is:
\[ (f \circ f)(x) = x^4 + 2x^2 + 2 \]

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