1. **Problem statement:** Find the Taylor series expansion of the function $f(x) = e^{2x}$ around $x=0$ up to the $x^3$ term. 2. **Formula:** The Taylor series of a function $f(x)$ around $x=a$ is given by: $$f(x) = \sum_{n=0}^\infty \frac{f^{(n)}(a)}{n!}(x - a)^n$$ where $f^{(n)}(a)$ is the $n$th derivative of $f$ evaluated at $x=a$. 3. Since we want the series around $x=0$, $a=0$, so: $$f(x) = \sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!} x^n$$ 4. **Calculate derivatives:** - $f(x) = e^{2x}$ - $f'(x) = 2e^{2x}$ - $f''(x) = 4e^{2x}$ - $f^{(3)}(x) = 8e^{2x}$ 5. **Evaluate derivatives at $x=0$:** - $f(0) = e^0 = 1$ - $f'(0) = 2e^0 = 2$ - $f''(0) = 4e^0 = 4$ - $f^{(3)}(0) = 8e^0 = 8$ 6. **Write the Taylor series up to $x^3$ term:** $$f(x) \approx f(0) + \frac{f'(0)}{1!}x + \frac{f''(0)}{2!}x^2 + \frac{f^{(3)}(0)}{3!}x^3$$ Substitute values: $$= 1 + 2x + \frac{4}{2}x^2 + \frac{8}{6}x^3 = 1 + 2x + 2x^2 + \frac{4}{3}x^3$$ 7. **Final answer:** $$\boxed{f(x) \approx 1 + 2x + 2x^2 + \frac{4}{3}x^3}$$ This is the Taylor series expansion of $e^{2x}$ around $x=0$ up to the $x^3$ term.