1. **State the problem:** We need to find the local extrema (maximum and minimum) of the function $$f(x) = 2xe^{-2x}$$ using the first derivative test. 2. **Find the first derivative:** Use the product rule: $$f'(x) = \frac{d}{dx}(2x) \cdot e^{-2x} + 2x \cdot \frac{d}{dx}(e^{-2x})$$ Calculate each part: $$\frac{d}{dx}(2x) = 2$$ $$\frac{d}{dx}(e^{-2x}) = -2e^{-2x}$$ So, $$f'(x) = 2e^{-2x} + 2x(-2e^{-2x}) = 2e^{-2x} - 4xe^{-2x} = 2e^{-2x}(1 - 2x)$$ 3. **Find critical points:** Set $$f'(x) = 0$$ Since $$2e^{-2x} \neq 0$$ for all real $$x$$, solve: $$1 - 2x = 0 \implies x = \frac{1}{2}$$ 4. **Use the first derivative test:** - For $$x < \frac{1}{2}$$, pick $$x=0$$: $$f'(0) = 2e^{0}(1 - 0) = 2 > 0$$ (positive) - For $$x > \frac{1}{2}$$, pick $$x=1$$: $$f'(1) = 2e^{-2}(1 - 2) = 2e^{-2}(-1) < 0$$ (negative) The derivative changes from positive to negative at $$x=\frac{1}{2}$$, so $$f$$ has a local maximum there. 5. **Find the function value at $$x=\frac{1}{2}$$:** $$f\left(\frac{1}{2}\right) = 2 \cdot \frac{1}{2} \cdot e^{-2 \cdot \frac{1}{2}} = 1 \cdot e^{-1} = e^{-1}$$ 6. **Check for local minimum:** There are no other critical points, so no local minimum. **Final answers:** - Local maximum at $$x=\frac{1}{2}$$ with value $$e^{-1}$$. - No local minimum.