Let f (x)= `x^3`-3x+1. Find the number of different real solution of the equation f (f(x) =0

Options

A. 2
B. 4
C. 5
D. 7

Explanation

To solve the problem of finding the number of different real solutions of the equation \( f(f(x)) = 0 \) where \( f(x) = x^3 - 3x + 1 \), we will follow these steps:

### Step 1: Analyze the function \( f(x) \)

The function is given by:
\[ f(x) = x^3 - 3x + 1 \]

### Step 2: Find the derivative \( f'(x) \)

To determine the monotonicity of \( f(x) \), we compute its derivative:
\[ f'(x) = 3x^2 - 3 = 3(x^2 - 1) = 3(x - 1)(x + 1) \]
This shows that \( f(x) \) has critical points at \( x = -1 \) and \( x = 1 \).

### Step 3: Determine the behavior of \( f(x) \)

- For \( x < -1 \), \( f'(x) > 0 \) (increasing).
- For \( -1 < x < 1 \), \( f'(x) < 0 \) (decreasing).
- For \( x > 1 \), \( f'(x) > 0 \) (increasing).

### Step 4: Evaluate \( f(x) \) at critical points and endpoints

Now we evaluate \( f(x) \) at the critical points and some additional points:
- \( f(-1) = (-1)^3 - 3(-1) + 1 = -1 + 3 + 1 = 3 \)
- \( f(0) = 0^3 - 3(0) + 1 = 1 \)
- \( f(1) = 1^3 - 3(1) + 1 = 1 - 3 + 1 = -1 \)
- \( f(3) = 3^3 - 3(3) + 1 = 27 - 9 + 1 = 19 \)

### Step 5: Identify the roots of \( f(x) = 0 \)

To find the roots of \( f(x) = 0 \), we note:
- \( f(-1) = 3 > 0 \)
- \( f(0) = 1 > 0 \)
- \( f(1) = -1 < 0 \)
- \( f(3) = 19 > 0 \)

By the Intermediate Value Theorem, there is at least one root in the interval \( (-1, 1) \) and at least one root in \( (1, 3) \).

### Step 6: Count the number of real roots of \( f(x) = 0 \)

Since \( f(x) \) is a cubic polynomial, it can have at most 3 real roots. Given the behavior of \( f(x) \), we conclude that:
- There are 3 real roots of \( f(x) = 0 \).

### Step 7: Analyze \( f(f(x)) = 0 \)

Now we need to find the solutions to \( f(f(x)) = 0 \). This means we need to solve:
1. \( f(x) = r_1 \)
2. \( f(x) = r_2 \)
3. \( f(x) = r_3 \)

Where \( r_1, r_2, r_3 \) are the three real roots of \( f(x) = 0 \).

### Step 8: Determine the number of solutions for each case

- For \( f(x) = r_1 \): Since \( r_1 \) is between \( -1 \) and \( 1 \), and \( f(x) \) is decreasing in that interval, it has 1 real root.
- For \( f(x) = r_2 \): Since \( r_2 \) is also between \( -1 \) and \( 1 \), it has 3 real roots.
- For \( f(x) = r_3 \): Since \( r_3 \) is greater than \( 1 \), it also has 3 real roots.

### Step 9: Total number of different real solutions

Adding these up:
- From \( f(x) = r_1 \): 1 root
- From \( f(x) = r_2 \): 3 roots
- From \( f(x) = r_3 \): 3 roots

Thus, the total number of different real solutions is:
\[ 1 + 3 + 3 = 7 \]

### Final Answer

The number of different real solutions of the equation \( f(f(x)) = 0 \) is **7**.

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