The numbr of real roots of the equation `x ^(2013)+ e ^(20144x) =0` is

Answer

To solve the equation \( x^{2013} + e^{20144x} = 0 \) and determine the number of real roots, we can follow these steps:

### Step 1: Define the function
Let \( f(x) = x^{2013} + e^{20144x} \).

### Step 2: Differentiate the function
We need to find the derivative \( f'(x) \) to analyze the behavior of the function:
\[ f'(x) = \frac{d}{dx}(x^{2013}) + \frac{d}{dx}(e^{20144x}) = 2013x^{2012} + 20144e^{20144x} \]

### Step 3: Analyze the derivative
Now, we analyze \( f'(x) \):
- The term \( 2013x^{2012} \) is always non-negative for all \( x \) because any even power of \( x \) is non-negative.
- The term \( 20144e^{20144x} \) is always positive for all \( x \) because the exponential function \( e^{20144x} \) is always positive.

Thus, we conclude that:
\[ f'(x) > 0 \quad \text{for all } x \]
This means that \( f(x) \) is a strictly increasing function.

### Step 4: Evaluate the function at extreme values
Next, we evaluate \( f(x) \) at \( x = -\infty \):
\[ \lim_{x \to -\infty} f(x) = \lim_{x \to -\infty} (x^{2013} + e^{20144x}) \]
- As \( x \to -\infty \), \( x^{2013} \to -\infty \) (since 2013 is odd).
- As \( x \to -\infty \), \( e^{20144x} \to 0 \).

Thus,
\[ \lim_{x \to -\infty} f(x) = -\infty \]

Now, we evaluate \( f(x) \) as \( x \to +\infty \):
\[ \lim_{x \to +\infty} f(x) = \lim_{x \to +\infty} (x^{2013} + e^{20144x}) \]
- As \( x \to +\infty \), both \( x^{2013} \) and \( e^{20144x} \) approach \( +\infty \).

Thus,
\[ \lim_{x \to +\infty} f(x) = +\infty \]

### Step 5: Conclusion about the roots
Since \( f(x) \) is a strictly increasing function that goes from \( -\infty \) to \( +\infty \), it must cross the x-axis exactly once. Therefore, the equation \( x^{2013} + e^{20144x} = 0 \) has exactly one real root.

### Final Answer
The number of real roots of the equation \( x^{2013} + e^{20144x} = 0 \) is **1**.