Let f(x)`=x^(13)+x^(11)+x^(9)+x^(7)+x^(5)+x^(3)+x+19`. The f(x) =0 has

Options

A. 13 real roots
B. only one positive and only two negative real roots
C. not more than one real root
D. has two positive and one negative real root

Answer

To determine how many real roots the equation \( f(x) = 0 \) has for the function \( f(x) = x^{13} + x^{11} + x^{9} + x^{7} + x^{5} + x^{3} + x + 19 \), we can use Descartes' Rule of Signs. Here’s a step-by-step solution:

### Step 1: Analyze \( f(x) \) for Positive Roots
1. **Write down the function**:
\[ f(x) = x^{13} + x^{11} + x^{9} + x^{7} + x^{5} + x^{3} + x + 19 \]

2. **Identify the signs of the coefficients**:
All coefficients of \( f(x) \) are positive (since all powers of \( x \) and the constant term 19 are positive).

3. **Count the number of sign changes**:
Since there are no sign changes in \( f(x) \), by Descartes' Rule of Signs, there are **0 positive real roots**.

### Step 2: Analyze \( f(-x) \) for Negative Roots
1. **Calculate \( f(-x) \)**:
\[ f(-x) = (-x)^{13} + (-x)^{11} + (-x)^{9} + (-x)^{7} + (-x)^{5} + (-x)^{3} + (-x) + 19 \]
This simplifies to:
\[ f(-x) = -x^{13} - x^{11} - x^{9} - x^{7} - x^{5} - x^{3} - x + 19 \]

2. **Identify the signs of the coefficients**:
The coefficients of the terms are:
- For \( -x^{13}, -x^{11}, -x^{9}, -x^{7}, -x^{5}, -x^{3}, -x \): all negative
- For \( +19 \): positive

3. **Count the number of sign changes**:
The sign changes from negative to positive occurs once (from the last negative term to the positive constant). Thus, there is **1 negative real root**.

### Step 3: Conclusion on the Number of Real Roots
- **Total Real Roots**:
- Positive real roots: 0
- Negative real roots: 1

Thus, the total number of real roots for \( f(x) = 0 \) is **1 real root**.

### Final Answer
The function \( f(x) = 0 \) has **1 real root**.

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