1. **State the problem:** We need to use the first derivative test to find the local extrema (maximum and minimum) of the function $$f(x) = 3x^2 - 4x$$ and determine their locations and values. 2. **Find the first derivative:** The first derivative of $$f(x)$$ is $$f'(x) = \frac{d}{dx}(3x^2 - 4x) = 6x - 4$$. 3. **Find critical points:** Set $$f'(x) = 0$$ to find critical points: $$6x - 4 = 0 \implies 6x = 4 \implies x = \frac{2}{3}$$. 4. **Use the first derivative test:** - For $$x < \frac{2}{3}$$, choose $$x=0$$: $$f'(0) = 6(0) - 4 = -4 < 0$$ (derivative negative, function decreasing). - For $$x > \frac{2}{3}$$, choose $$x=1$$: $$f'(1) = 6(1) - 4 = 2 > 0$$ (derivative positive, function increasing). Since $$f'(x)$$ changes from negative to positive at $$x=\frac{2}{3}$$, the function has a local minimum there. 5. **Find the function value at the minimum:** $$f\left(\frac{2}{3}\right) = 3\left(\frac{2}{3}\right)^2 - 4\left(\frac{2}{3}\right) = 3\left(\frac{4}{9}\right) - \frac{8}{3} = \frac{12}{9} - \frac{8}{3} = \frac{4}{3} - \frac{8}{3} = -\frac{4}{3}$$. 6. **Check for local maximum:** Since the derivative does not change from positive to negative at any critical point, there is no local maximum. **Final answers:** - Local minimum of $$-\frac{4}{3}$$ at $$x=\frac{2}{3}$$. - No local maximum.