1. **State the problem:** We are given the function $f(x) = 3x e^{-4x}$ and need to find the location and value of any local maxima and minima. 2. **Recall the formula and rules:** To find local maxima and minima, we first find the critical points by setting the derivative $f'(x)$ equal to zero. Then, we use the second derivative test or analyze the sign changes of $f'(x)$ to classify these points. 3. **Find the first derivative:** Using the product rule for $f(x) = u(x)v(x)$ where $u=3x$ and $v=e^{-4x}$, $$f'(x) = u'v + uv' = 3 e^{-4x} + 3x (-4) e^{-4x} = 3 e^{-4x} - 12x e^{-4x} = e^{-4x}(3 - 12x)$$ 4. **Find critical points:** Set $f'(x) = 0$: $$e^{-4x}(3 - 12x) = 0$$ Since $e^{-4x} \neq 0$ for all real $x$, we solve: $$3 - 12x = 0 \implies 12x = 3 \implies x = \frac{1}{4}$$ 5. **Find the second derivative:** Differentiate $f'(x) = e^{-4x}(3 - 12x)$ using product rule: $$f''(x) = \frac{d}{dx}[e^{-4x}](3 - 12x) + e^{-4x} \frac{d}{dx}(3 - 12x)$$ $$= (-4 e^{-4x})(3 - 12x) + e^{-4x}(-12) = e^{-4x}[-4(3 - 12x) - 12] = e^{-4x}(-12 + 48x - 12) = e^{-4x}(48x - 24)$$ 6. **Evaluate $f''(x)$ at $x=\frac{1}{4}$:** $$f''\left(\frac{1}{4}\right) = e^{-4(\frac{1}{4})}(48 \cdot \frac{1}{4} - 24) = e^{-1}(12 - 24) = e^{-1}(-12) < 0$$ Since $f''(\frac{1}{4}) < 0$, the function has a local maximum at $x=\frac{1}{4}$. 7. **Find the function value at $x=\frac{1}{4}$:** $$f\left(\frac{1}{4}\right) = 3 \cdot \frac{1}{4} \cdot e^{-4 \cdot \frac{1}{4}} = \frac{3}{4} e^{-1}$$ 8. **Check for local minima:** There are no other critical points, and the function tends to zero as $x \to \infty$ and $x \to 0^+$, so no local minimum exists. **Final answers:** - Local maximum of $\frac{3}{4} e^{-1}$ at $x=\frac{1}{4}$. - No local minimum.