1. **State the problem:** Find the second derivative and determine if the function $f(x) = xe^{x/2}$ has any local maxima or minima. 2. **Recall the formulas:** - The first derivative $f'(x)$ gives the slope of the function. - The second derivative $f''(x)$ tells us about the concavity. - Critical points occur where $f'(x) = 0$. - If $f''(x) > 0$ at a critical point, it's a local minimum. - If $f''(x) < 0$ at a critical point, it's a local maximum. 3. **Find the first derivative $f'(x)$:** Use the product rule: $\frac{d}{dx}[u v] = u' v + u v'$ where $u = x$ and $v = e^{x/2}$. Calculate: $$u' = 1$$ $$v' = e^{x/2} \cdot \frac{1}{2} = \frac{1}{2} e^{x/2}$$ So, $$f'(x) = 1 \cdot e^{x/2} + x \cdot \frac{1}{2} e^{x/2} = e^{x/2} + \frac{x}{2} e^{x/2} = e^{x/2} \left(1 + \frac{x}{2}\right)$$ 4. **Find critical points by setting $f'(x) = 0$:** Since $e^{x/2} > 0$ for all real $x$, set the factor in parentheses to zero: $$1 + \frac{x}{2} = 0 \implies x = -2$$ 5. **Find the second derivative $f''(x)$:** Differentiate $f'(x) = e^{x/2} \left(1 + \frac{x}{2}\right)$ using the product rule again. Let $u = e^{x/2}$ and $w = 1 + \frac{x}{2}$. Calculate: $$u' = \frac{1}{2} e^{x/2}$$ $$w' = \frac{1}{2}$$ Then, $$f''(x) = u' w + u w' = \frac{1}{2} e^{x/2} \left(1 + \frac{x}{2}\right) + e^{x/2} \cdot \frac{1}{2} = e^{x/2} \left( \frac{1}{2} + \frac{x}{4} + \frac{1}{2} \right) = e^{x/2} \left(1 + \frac{x}{4}\right)$$ 6. **Evaluate $f''(x)$ at the critical point $x = -2$:** $$f''(-2) = e^{-2/2} \left(1 + \frac{-2}{4}\right) = e^{-1} \left(1 - \frac{1}{2}\right) = e^{-1} \cdot \frac{1}{2} > 0$$ Since $f''(-2) > 0$, the function has a local minimum at $x = -2$. **Final answer:** - The second derivative is $$f''(x) = e^{x/2} \left(1 + \frac{x}{4}\right)$$ - There is a local minimum at $x = -2$.