1. **Problem statement:** Consider the function $f(x) = 2x^2 - 3x + 1$. We need to find: 1) The domain of definition of $f(x)$. 2) The limits of $f(x)$ at the open boundaries of its domain. 3) The derivative of $f(x)$. 2. **Domain of definition:** The function $f(x) = 2x^2 - 3x + 1$ is a polynomial. Important rule: Polynomials are defined for all real numbers. Therefore, the domain is all real numbers: $$\text{Domain} = (-\infty, +\infty)$$ 3. **Limits at the boundaries:** Since the domain is all real numbers, the boundaries are $x \to -\infty$ and $x \to +\infty$. Calculate $$\lim_{x \to -\infty} f(x) = \lim_{x \to -\infty} (2x^2 - 3x + 1)$$ The term $2x^2$ dominates as $x \to -\infty$, and since $x^2$ is positive and grows without bound, the limit is: $$\lim_{x \to -\infty} f(x) = +\infty$$ Similarly, calculate $$\lim_{x \to +\infty} f(x) = \lim_{x \to +\infty} (2x^2 - 3x + 1) = +\infty$$ 4. **Derivative calculation:** Use the power rule: $$\frac{d}{dx} x^n = n x^{n-1}$$ Derivative of $f(x)$: $$f'(x) = \frac{d}{dx} (2x^2) - \frac{d}{dx} (3x) + \frac{d}{dx} (1) = 2 \cdot 2x^{2-1} - 3 \cdot 1 + 0 = 4x - 3$$ **Final answers:** - Domain: $(-\infty, +\infty)$ - Limits: $\lim_{x \to -\infty} f(x) = +\infty$, $\lim_{x \to +\infty} f(x) = +\infty$ - Derivative: $f'(x) = 4x - 3$