1. Stating the problem: We want to determine in which interval the function $f(x) = 3 - 2e^{3x^2}$ is decreasing. 2. First, we find the derivative $f'(x)$ to analyze where $f$ is increasing or decreasing. $$f'(x) = \frac{d}{dx} \left(3 - 2e^{3x^2}\right) = -2 \cdot \frac{d}{dx} \left(e^{3x^2}\right)$$ 3. Using the chain rule, $$\frac{d}{dx} e^{3x^2} = e^{3x^2} \cdot \frac{d}{dx} (3x^2) = e^{3x^2} \cdot 6x$$ Hence, $$f'(x) = -2 \cdot e^{3x^2} \cdot 6x = -12x e^{3x^2}$$ 4. Since $e^{3x^2}$ is always positive for all real $x$, the sign of $f'(x)$ depends on the sign of $-12x$. - For $x > 0$: $-12x < 0$, so $f'(x) < 0$, meaning $f$ is decreasing. - For $x < 0$: $-12x > 0$, so $f'(x) > 0$, meaning $f$ is increasing. 5. Therefore, the function $f$ is decreasing on the interval $]0, \infty[$. Final answer: The function is decreasing on interval (c) $]0, \infty[$.