1. **Problem statement:** Find the maximum and minimum values of the function $$f(x) = \frac{2x^2 + 4x + 5}{x^2 + 1}$$ for all real numbers $x \in \mathbb{R}$. 2. **Formula and rules:** To find extrema of a function, we first find its derivative and set it equal to zero to find critical points. Then, we analyze these points to determine if they are maxima, minima, or neither. 3. **Calculate the derivative:** Using the quotient rule: $$f'(x) = \frac{(4x + 4)(x^2 + 1) - (2x^2 + 4x + 5)(2x)}{(x^2 + 1)^2}$$ 4. **Simplify the numerator:** $$\begin{aligned} & (4x + 4)(x^2 + 1) - (2x^2 + 4x + 5)(2x) \\ =& (4x^3 + 4x + 4x^2 + 4) - (4x^3 + 8x^2 + 10x) \\ =& 4x^3 + 4x + 4x^2 + 4 - 4x^3 - 8x^2 - 10x \\ =& (4x^3 - 4x^3) + (4x^2 - 8x^2) + (4x - 10x) + 4 \\ =& -4x^2 - 6x + 4 \end{aligned}$$ 5. **Set the numerator equal to zero to find critical points:** $$-4x^2 - 6x + 4 = 0$$ Multiply both sides by $-1$: $$4x^2 + 6x - 4 = 0$$ 6. **Simplify by dividing by 2:** $$\cancel{2} \times (2x^2 + 3x - 2) = 0 \Rightarrow 2x^2 + 3x - 2 = 0$$ 7. **Solve quadratic equation:** $$x = \frac{-3 \pm \sqrt{3^2 - 4 \times 2 \times (-2)}}{2 \times 2} = \frac{-3 \pm \sqrt{9 + 16}}{4} = \frac{-3 \pm 5}{4}$$ 8. **Find roots:** $$x_1 = \frac{-3 + 5}{4} = \frac{2}{4} = 0.5$$ $$x_2 = \frac{-3 - 5}{4} = \frac{-8}{4} = -2$$ 9. **Evaluate $f(x)$ at critical points:** $$f(0.5) = \frac{2(0.5)^2 + 4(0.5) + 5}{(0.5)^2 + 1} = \frac{2(0.25) + 2 + 5}{0.25 + 1} = \frac{0.5 + 2 + 5}{1.25} = \frac{7.5}{1.25} = 6$$ $$f(-2) = \frac{2(4) + 4(-2) + 5}{4 + 1} = \frac{8 - 8 + 5}{5} = \frac{5}{5} = 1$$ 10. **Check horizontal asymptote:** As $x \to \pm \infty$, $$f(x) \to \frac{2x^2}{x^2} = 2$$ 11. **Determine max and min:** - At $x=0.5$, $f(0.5) = 6$ is a local maximum. - At $x=-2$, $f(-2) = 1$ is a local minimum. - The horizontal asymptote $y=2$ is the end behavior. **Final answer:** $$\boxed{\text{Maximum value } = 6 \text{ at } x=0.5, \quad \text{Minimum value } = 1 \text{ at } x=-2}$$