1. **Statement of the problem:** We have a piecewise function defined as: $$f(x) = \begin{cases} e^x - 1 & x > 0 \\ \ln(x + 1) & x = 0 \\ x^3 e^{2x} & x < 0 \end{cases}$$ We need to: - Find the domain of $f$. - Check continuity of $f$ at $x=0$. - Determine if $f$ is differentiable at $x=0$. - Calculate the derivative $f'(x)$ for all $x$. 2. **Domain of $f$:** - For $x > 0$, $f(x) = e^x - 1$ is defined for all real $x$. - At $x=0$, $f(0) = \ln(0 + 1) = \ln(1) = 0$. - For $x < 0$, $f(x) = x^3 e^{2x}$ is defined for all real $x$. Therefore, the domain of $f$ is all real numbers $\mathbb{R}$. 3. **Continuity at $x=0$:** We check if $\lim_{x \to 0^-} f(x) = f(0) = \lim_{x \to 0^+} f(x)$. - Left limit: $$\lim_{x \to 0^-} x^3 e^{2x} = 0^3 \cdot e^0 = 0$$ - Right limit: $$\lim_{x \to 0^+} e^x - 1 = e^0 - 1 = 0$$ - Value at zero: $$f(0) = \ln(1) = 0$$ Since all three are equal to 0, $f$ is continuous at $x=0$. 4. **Differentiability at $x=0$:** We check if the left and right derivatives at 0 are equal. - For $x < 0$, $$f(x) = x^3 e^{2x}$$ Using product rule: $$f'(x) = 3x^2 e^{2x} + x^3 \cdot 2 e^{2x} = e^{2x} (3x^2 + 2x^3)$$ Evaluate left derivative at 0: $$f'_-(0) = e^0 (3 \cdot 0^2 + 2 \cdot 0^3) = 0$$ - For $x > 0$, $$f(x) = e^x - 1$$ Derivative: $$f'(x) = e^x$$ Evaluate right derivative at 0: $$f'_+(0) = e^0 = 1$$ Since $f'_-(0) = 0 \neq 1 = f'_+(0)$, $f$ is not differentiable at $x=0$. 5. **Derivative of $f$ for all $x$:** - For $x < 0$: $$f'(x) = e^{2x} (3x^2 + 2x^3)$$ - For $x > 0$: $$f'(x) = e^x$$ - At $x=0$, derivative does not exist. - For $x=0$, derivative from left and right differ, so no derivative at 0. **Final answers:** - Domain: $\mathbb{R}$ - $f$ is continuous at $x=0$. - $f$ is not differentiable at $x=0$. - Derivative: $$f'(x) = \begin{cases} e^{2x} (3x^2 + 2x^3) & x < 0 \\ \text{undefined} & x = 0 \\ e^x & x > 0 \end{cases}$$