1. **State the problem:** We are given the function $g(x) = x^2 e^{4x}$ and need to find intervals of concavity, points of inflection, local extrema, and determine if there are any global extrema. 2. **Find the first derivative $g'(x)$:** Use the product rule $\frac{d}{dx}[u v] = u'v + uv'$ where $u = x^2$ and $v = e^{4x}$. $$u' = 2x, \quad v' = 4 e^{4x}$$ So, $$g'(x) = 2x e^{4x} + x^2 (4 e^{4x}) = e^{4x}(2x + 4x^2) = 2x e^{4x}(1 + 2x)$$ 3. **Find the second derivative $g''(x)$:** Differentiate $g'(x)$ again using product rule on $2x e^{4x}(1 + 2x)$. Let $h(x) = 2x(1 + 2x) = 2x + 4x^2$, then $$g'(x) = e^{4x} h(x)$$ Then, $$g''(x) = e^{4x} h'(x) + 4 e^{4x} h(x) = e^{4x} (h'(x) + 4 h(x))$$ Calculate $h'(x)$: $$h'(x) = 2 + 8x$$ So, $$g''(x) = e^{4x} ((2 + 8x) + 4(2x + 4x^2)) = e^{4x} (2 + 8x + 8x + 16x^2) = e^{4x} (2 + 16x + 16x^2)$$ Simplify: $$g''(x) = e^{4x} (16x^2 + 16x + 2)$$ 4. **Determine concavity:** Since $e^{4x} > 0$ for all $x$, the sign of $g''(x)$ depends on the quadratic $16x^2 + 16x + 2$. Divide by 2 for simplicity: $$8x^2 + 8x + 1$$ Find roots using quadratic formula: $$x = \frac{-8 \pm \sqrt{64 - 32}}{16} = \frac{-8 \pm \sqrt{32}}{16} = \frac{-8 \pm 4\sqrt{2}}{16} = \frac{-2 \pm \sqrt{2}}{4}$$ 5. **Intervals of concavity:** - Concave up where $g''(x) > 0$: $(-\infty, \frac{-2 - \sqrt{2}}{4}]$ and $[\frac{-2 + \sqrt{2}}{4}, \infty)$ - Concave down where $g''(x) < 0$: $[\frac{-2 - \sqrt{2}}{4}, \frac{-2 + \sqrt{2}}{4}]$ 6. **Points of inflection:** Occur where $g''(x) = 0$, i.e., at $$x = \frac{-2 - \sqrt{2}}{4}, \quad x = \frac{-2 + \sqrt{2}}{4}$$ Evaluate $g(x)$ at these points for coordinates. 7. **Find local extrema:** Set $g'(x) = 0$: $$2x e^{4x} (1 + 2x) = 0$$ Since $e^{4x} \neq 0$, solve: $$2x = 0 \Rightarrow x=0$$ $$1 + 2x = 0 \Rightarrow x = -\frac{1}{2}$$ Evaluate $g(x)$ at these points: $$g(0) = 0^2 e^0 = 0$$ $$g(-\frac{1}{2}) = \left(-\frac{1}{2}\right)^2 e^{4(-1/2)} = \frac{1}{4} e^{-2}$$ Determine nature using second derivative test: $$g''(0) = e^0 (16(0)^2 + 16(0) + 2) = 2 > 0 \Rightarrow \text{local minimum at } x=0$$ $$g''(-\frac{1}{2}) = e^{-2} (16(\frac{1}{4}) - 8 + 2) = e^{-2} (4 - 8 + 2) = e^{-2} (-2) < 0 \Rightarrow \text{local maximum at } x = -\frac{1}{2}$$ 8. **Global extrema:** Since $g(x) = x^2 e^{4x}$ grows very large as $x \to \infty$ and tends to 0 as $x \to -\infty$, there is no global maximum (function grows without bound), but there is a global minimum at $x=0$ with value 0. **Final answers:** - Concave up on $(-\infty, \frac{-2 - \sqrt{2}}{4}]$, $[\frac{-2 + \sqrt{2}}{4}, \infty)$ - Concave down on $[\frac{-2 - \sqrt{2}}{4}, \frac{-2 + \sqrt{2}}{4}]$ - Points of inflection at $x = \frac{-2 - \sqrt{2}}{4}, \frac{-2 + \sqrt{2}}{4}$ with coordinates $\left(\frac{-2 - \sqrt{2}}{4}, g\left(\frac{-2 - \sqrt{2}}{4}\right)\right)$ and $\left(\frac{-2 + \sqrt{2}}{4}, g\left(\frac{-2 + \sqrt{2}}{4}\right)\right)$ - Local maximum at $\left(-\frac{1}{2}, \frac{1}{4} e^{-2}\right)$ - Local minimum at $(0,0)$ - Global minimum at $(0,0)$, no global maximum.