The numbr of real roots of the equation `x ^(2013)+ e ^(2014x) =0` is

Answer

To find the number of real roots of the equation \( x^{2013} + e^{2014x} = 0 \), we can follow these steps:

### Step 1: Define the function
Let \( y = f(x) = x^{2013} + e^{2014x} \).

### Step 2: Differentiate the function
We differentiate \( f(x) \) with respect to \( x \):
\[ f'(x) = \frac{dy}{dx} = 2013x^{2012} + 2014e^{2014x} \]

### Step 3: Analyze the derivative
Now, we analyze the derivative \( f'(x) \):
- The term \( 2013x^{2012} \) is always non-negative because \( x^{2012} \) is a non-negative quantity (as it is raised to an even power).
- The term \( 2014e^{2014x} \) is always positive for all real \( x \) since the exponential function \( e^{2014x} \) is always positive.

Thus, we conclude that:
\[ f'(x) = 2013x^{2012} + 2014e^{2014x} > 0 \quad \text{for all } x \]
This means that \( f'(x) \) is strictly positive for all \( x \).

### Step 4: Determine the behavior of the function
Since \( f'(x) > 0 \) for all \( x \), the function \( f(x) \) is strictly increasing. A strictly increasing function can cross the x-axis at most once.

### Step 5: Evaluate the limits
Now, we evaluate the limits of \( f(x) \) as \( x \) approaches negative and positive infinity:
- As \( x \to -\infty \), \( x^{2013} \to -\infty \) and \( e^{2014x} \to 0 \), thus \( f(x) \to -\infty \).
- As \( x \to +\infty \), \( x^{2013} \to +\infty \) and \( e^{2014x} \to +\infty \), thus \( f(x) \to +\infty \).

### Step 6: Conclusion on the number of roots
Since \( f(x) \) is strictly increasing and transitions from \( -\infty \) to \( +\infty \), it must cross the x-axis exactly once.

Therefore, the number of real roots of the equation \( x^{2013} + e^{2014x} = 0 \) is:
\[ \boxed{1} \]