1. **State the problem:** We are given the function $$f(x) = \frac{x^4 + 4x^3 + 6x^2 + 4x + 1}{x^2 + 2x + 1}$$ and need to find the values of the derivative $f'(x)$ at $x = -2, -1, 0$ and the function values $f(x)$ at the same points. 2. **Simplify the function:** Notice the numerator is a binomial expansion: $$x^4 + 4x^3 + 6x^2 + 4x + 1 = (x+1)^4$$ The denominator is: $$x^2 + 2x + 1 = (x+1)^2$$ So, $$f(x) = \frac{(x+1)^4}{(x+1)^2}$$ 3. **Simplify the fraction:** Cancel common factors: $$f(x) = \frac{\cancel{(x+1)^2} \cdot (x+1)^2}{\cancel{(x+1)^2}} = (x+1)^2$$ 4. **Find the derivative:** Using the power rule: $$f'(x) = 2(x+1)$$ 5. **Evaluate $f(x)$ and $f'(x)$ at the points:** - At $x = -2$: $$f(-2) = (-2 + 1)^2 = (-1)^2 = 1$$ $$f'(-2) = 2(-2 + 1) = 2(-1) = -2$$ - At $x = -1$: $$f(-1) = (-1 + 1)^2 = 0^2 = 0$$ $$f'(-1) = 2(-1 + 1) = 2(0) = 0$$ - At $x = 0$: $$f(0) = (0 + 1)^2 = 1^2 = 1$$ $$f'(0) = 2(0 + 1) = 2(1) = 2$$ **Final answers:** $$f'(-2) = -2, \quad f'(-1) = 0, \quad f'(0) = 2$$ $$f(-2) = 1, \quad f(-1) = 0, \quad f(0) = 1$$