In the function f(x) = (x+a)/(sqrt(x²+b)), determine

In the function $\displaystyle f(x) = \frac{x+a}{\sqrt{x^{2}+b}}$, determine the parameters $a$ and $b$ so that $f$ has an extremum at $x = -1$ and an inflection point at $x = -2$.

Step 1: The first derivative

With $u = x+a$ and $v = \bigl(x^{2}+b\bigr)^{1/2}$, so that $v' = \dfrac{x}{\sqrt{x^{2}+b}}$: \[f'(x) = \frac{\sqrt{x^{2}+b}-\bigl(x+a\bigr)\dfrac{x}{\sqrt{x^{2}+b}}}{x^{2}+b} = \frac{\bigl(x^{2}+b\bigr)-x\bigl(x+a\bigr)}{\bigl(x^{2}+b\bigr)^{3/2}} = \frac{b-ax}{\bigl(x^{2}+b\bigr)^{3/2}}. \tag{1}\]

Step 2: The extremum condition

A necessary condition for an extremum at $x = -1$ is $f'(-1) = 0$: \[f'(-1) = \frac{b+a}{\bigl(1+b\bigr)^{3/2}} = 0 \;\Longrightarrow\; a+b = 0 \;\Longrightarrow\; a = -b. \tag{2}\]

Step 3: The second derivative

Differentiating (1) with $P = b-ax$ and $Q = \bigl(x^{2}+b\bigr)^{3/2}$, so $P' = -a$ and $Q' = 3x\bigl(x^{2}+b\bigr)^{1/2}$, then cancelling one factor $\bigl(x^{2}+b\bigr)^{1/2}$:

\[ f''(x) = \frac{-a\bigl(x^{2}+b\bigr)-3x\bigl(b-ax\bigr)}{\bigl(x^{2}+b\bigr)^{5/2}} = \frac{2ax^{2}-3bx-ab}{\bigl(x^{2}+b\bigr)^{5/2}}. \tag{3} \]

Step 4: The inflection condition

A necessary condition for an inflection at $x = -2$ is $f''(-2) = 0$: \[\frac{8a+6b-ab}{\bigl(4+b\bigr)^{5/2}} = 0 \;\Longrightarrow\; 8a+6b-ab = 0. \tag{4}\]

Step 5: Solve the system

Substituting $a = -b$ from (2) into (4): \[-8b+6b+b^{2} = 0 \;\Longrightarrow\; b^{2}-2b = 0 \;\Longrightarrow\; b\bigl(b-2\bigr) = 0.\]

The root $b = 0$ must be rejected: it gives $a = 0$ and $f(x) = \dfrac{x}{\sqrt{x^{2}}} = \operatorname{sgn}x$, a step function with neither extremum nor inflection. The remaining root gives

\[ b = 2, \qquad a = -2. \]

Step 6: Verify the conditions are really met

With these values $f(x) = \dfrac{x-2}{\sqrt{x^{2}+2}}$ and, from (1), \[f'(x) = \frac{2+2x}{\bigl(x^{2}+2\bigr)^{3/2}} = \frac{2\bigl(x+1\bigr)}{\bigl(x^{2}+2\bigr)^{3/2}},\] which is negative for $x\lt -1$ and positive for $x\gt -1$: $x = -1$ is a genuine local minimum, with $f(-1) = -\sqrt3 = -1.7320508$. From (3),

\[ f''(x) = \frac{-4x^{2}-6x+4}{\bigl(x^{2}+2\bigr)^{5/2}} = \frac{-2\bigl(2x-1\bigr)\bigl(x+2\bigr)}{\bigl(x^{2}+2\bigr)^{5/2}}, \] so $f''\gt 0$ for $x\lt -2$ and $f''\lt 0$ for $-2\lt x\lt \tfrac12$: the concavity really does change at $x = -2$, which is therefore a genuine inflection point.

Step 7: Verification

\[f(-2) = \frac{-4}{\sqrt6} = -1.6329932; \qquad f(-1) = -\sqrt3 = -1.7320508; \qquad f(0) = -\sqrt2 = -1.4142136 \;\;\checkmark\]

\[\text{The second zero of } f'' \text{ is } x = \tfrac12,\ \text{a second inflection point} \;\;\checkmark\]

\[f\to\pm1 \text{ as } x\to\pm\infty,\ \text{so the minimum } -\sqrt3 \text{ is the global one} \;\;\checkmark\]