How many real solutions does the equation `x^7+14 x^5+16 x^3+30 x-560=0` have?

Options

A. 1
B. 3
C. 5
D. 7

Explanation

To determine how many real solutions the equation \( x^7 + 14x^5 + 16x^3 + 30x - 560 = 0 \) has, we can follow these steps:

### Step 1: Define the function
Let \( f(x) = x^7 + 14x^5 + 16x^3 + 30x - 560 \).

### Step 2: Find the derivative
To analyze the behavior of the function, we need to find its derivative:
\[ f'(x) = \frac{d}{dx}(x^7 + 14x^5 + 16x^3 + 30x - 560) \]
Calculating the derivative:
\[ f'(x) = 7x^6 + 70x^4 + 48x^2 + 30 \]

### Step 3: Analyze the derivative
Notice that \( f'(x) \) is a sum of non-negative terms:
- \( 7x^6 \) is non-negative for all \( x \).
- \( 70x^4 \) is non-negative for all \( x \).
- \( 48x^2 \) is non-negative for all \( x \).
- \( 30 \) is a positive constant.

Since all terms in \( f'(x) \) are non-negative, we conclude that:
\[ f'(x) > 0 \quad \text{for all } x \in \mathbb{R} \]
This means that \( f(x) \) is a strictly increasing function.

### Step 4: Analyze the limits of the function
Next, we check the behavior of \( f(x) \) as \( x \) approaches infinity and negative infinity:
- As \( x \to \infty \), \( f(x) \to \infty \) (since the leading term \( x^7 \) dominates).
- As \( x \to -\infty \), \( f(x) \to -\infty \) (the leading term \( x^7 \) also dominates but goes to negative infinity).

### Step 5: Apply the Intermediate Value Theorem
Since \( f(x) \) is strictly increasing, it can cross the x-axis at most once. Given that \( f(x) \) approaches \( -\infty \) as \( x \to -\infty \) and \( +\infty \) as \( x \to +\infty \), there must be exactly one point where \( f(x) = 0 \).

### Conclusion
Thus, the equation \( x^7 + 14x^5 + 16x^3 + 30x - 560 = 0 \) has exactly **one real solution**.

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