Consider the polynomial `f(x)=1 + 2x + 3x^2 +4x^3` for all `x in R` So f(x) has exactly one real root in the interval

Options

A. `(-1//4,0)`
B. `(-11 ,-3 //4)`
C. `(-3 //4, -1//2)`
D. `(0 ,1//4)`

Explanation

To find the interval in which the polynomial \( f(x) = 1 + 2x + 3x^2 + 4x^3 \) has exactly one real root, we can follow these steps:

### Step 1: Differentiate the Function
First, we need to find the derivative of the function \( f(x) \):
\[ f'(x) = \frac{d}{dx}(1 + 2x + 3x^2 + 4x^3) = 2 + 6x + 12x^2 \]

### Step 2: Analyze the Derivative
Next, we need to analyze the derivative \( f'(x) \) to determine the behavior of the function \( f(x) \):
\[ f'(x) = 12x^2 + 6x + 2 \]
This is a quadratic function. To find its roots, we can use the discriminant:
\[ D = b^2 - 4ac = (6)^2 - 4(12)(2) = 36 - 96 = -60 \]
Since the discriminant \( D \) is negative, the quadratic \( f'(x) \) has no real roots.

### Step 3: Determine the Sign of the Derivative
Since the coefficient of \( x^2 \) in \( f'(x) \) is positive (12), the derivative \( f'(x) \) is always positive for all \( x \in \mathbb{R} \). This means that \( f(x) \) is an increasing function.

### Step 4: Analyze the Behavior of the Function
Since \( f(x) \) is an increasing function, it can cross the x-axis at most once. Therefore, \( f(x) \) has exactly one real root.

### Step 5: Find the Interval for the Root
To find the interval where the root lies, we can evaluate \( f(x) \) at specific points:
1. Evaluate \( f(-1) \):
\[ f(-1) = 1 + 2(-1) + 3(-1)^2 + 4(-1)^3 = 1 - 2 + 3 - 4 = -2 \quad (\text{negative}) \]
2. Evaluate \( f(0) \):
\[ f(0) = 1 + 2(0) + 3(0)^2 + 4(0)^3 = 1 \quad (\text{positive}) \]

Since \( f(-1) < 0 \) and \( f(0) > 0 \), by the Intermediate Value Theorem, there is a root in the interval \( (-1, 0) \).

### Conclusion
Thus, the polynomial \( f(x) = 1 + 2x + 3x^2 + 4x^3 \) has exactly one real root in the interval \( (-1, 0) \).

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