Find the minimum and maximum values of f(x) = x/(x²+2x+4)

Find the minimum and maximum values of $\displaystyle f(x) = \frac{x}{x^{2}+2x+4}$.

Step 1: Domain

Completing the square, $x^{2}+2x+4 = \bigl(x+1\bigr)^{2}+3\ge3\gt 0$, so the denominator never vanishes and the domain is all of $\mathbb{R}$.

Step 2: The derivative

\[ f'(x) = \frac{\bigl(x^{2}+2x+4\bigr)-x\bigl(2x+2\bigr)}{\bigl(x^{2}+2x+4\bigr)^{2}} = \frac{4-x^{2}}{\bigl(x^{2}+2x+4\bigr)^{2}} = \frac{-\bigl(x-2\bigr)\bigl(x+2\bigr)}{\bigl(x^{2}+2x+4\bigr)^{2}}. \tag{1} \]

Step 3: Critical points and signs

The denominator is positive, so $f'$ vanishes only where $x = \pm2$, and its sign is that of $4-x^{2}$: \[x\lt -2:\ f'\lt 0\ (\searrow); \qquad -2\lt x\lt 2:\ f'\gt 0\ (\nearrow); \qquad x\gt 2:\ f'\lt 0\ (\searrow).\]

Step 4: The two extreme values

\[ x = -2:\ -\to+\ \text{minimum}, \quad f(-2) = \frac{-2}{4-4+4} = -\frac12; \]

\[ x = 2:\ +\to-\ \text{maximum}, \quad f(2) = \frac{2}{4+4+4} = \frac16. \]

Step 5: Why they are global

Since the degree of the denominator exceeds that of the numerator, $\lim_{x\to\pm\infty}f(x) = 0$.

The function decreases to $-\tfrac12$, rises to $\tfrac16$, then decreases back towards $0$. As $-\tfrac12\lt 0\lt \tfrac16$, no value outside $\left[-\tfrac12,\tfrac16\right]$ is ever attained: both local extrema are global.

Step 6: Verification

\[f(-4) = -\frac13 = -0.3333 \gt f(-2) = -0.5 \lt f(0) = 0 \;\;\checkmark\]

\[f(0) = 0 \lt f(2) = 0.1667 \gt f(4) = \frac17 = 0.1429 \;\;\checkmark\]

\[\text{Solving } y = \frac{x}{x^{2}+2x+4} \text{ for } x:\ yx^{2}+\bigl(2y-1\bigr)x+4y = 0 \text{ has real roots iff } -\tfrac12\le y\le\tfrac16 \;\;\checkmark\]