For small values of $x$, we can use a Taylor series expansion to approximate $x + e^x - e^{3x}$.

1. Expand $e^x$ around $x = 0$:

   The Taylor series for $e^x$ is: $e^x \approx 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \dots$

   Similarly, the Taylor series for $e^{3x}$ is: $e^{3x} \approx 1 + 3x + \frac{(3x)^2}{2} + \frac{(3x)^3}{6} + \dots$

2. Apply these approximations in $x + e^x - e^{3x}$:

   Substitute the expansions: $x + e^x - e^{3x} \approx x + \left(1 + x + \frac{x^2}{2}\right) - \left(1 + 3x + \frac{9x^2}{2}\right)$

3. Simplify the expression:

   $x + e^x - e^{3x} \approx x + 1 + x + \frac{x^2}{2} - 1 - 3x - \frac{9x^2}{2}$ $= -x - 4x^2$

4. Result:

   For small $x$, a good approximation is: $x + e^x - e^{3x} \approx -x - 4x^2$

Would you like a more detailed breakdown or further explanation?

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1. How does the Taylor series work for different functions?
2. Why does Taylor series approximation work better for smaller values?
3. Can higher-order terms improve this approximation?
4. How does this approximation change with larger $x$?
5. How can we test the accuracy of this approximation?

Tip: Taylor series approximations are often more accurate the closer the point is to the center of expansion (e.g., $x = 0$ here).