Derivative Critical Points 37D5Ea

1. **State the problem:** We are given the function $$f(x) = x^3 - 7x^2$$ and need to find its derivative $$f'(x)$$, solve $$f'(x) = 0$$, and find the points on the graph where the tangent line is horizontal. 2. **Find the derivative:** The derivative of $$f(x)$$ is found using the power rule: $$f'(x) = \frac{d}{dx}(x^3) - \frac{d}{dx}(7x^2) = 3x^2 - 14x$$ 3. **Solve $$f'(x) = 0$$:** Set the derivative equal to zero: $$3x^2 - 14x = 0$$ Factor out $$x$$: $$x(3x - 14) = 0$$ This gives two solutions: $$x = 0$$ or $$3x - 14 = 0$$ Solve the second: $$3x = 14$$ $$x = \frac{14}{3}$$ 4. **Find points where tangent is horizontal:** Substitute each $$x$$ back into $$f(x)$$: - For $$x=0$$: $$f(0) = 0^3 - 7(0)^2 = 0$$ - For $$x=\frac{14}{3}$$: $$f\left(\frac{14}{3}\right) = \left(\frac{14}{3}\right)^3 - 7\left(\frac{14}{3}\right)^2$$ Calculate stepwise: $$\left(\frac{14}{3}\right)^3 = \frac{14^3}{3^3} = \frac{2744}{27}$$ $$\left(\frac{14}{3}\right)^2 = \frac{196}{9}$$ So, $$f\left(\frac{14}{3}\right) = \frac{2744}{27} - 7 \times \frac{196}{9} = \frac{2744}{27} - \frac{1372}{9}$$ Convert to common denominator 27: $$\frac{1372}{9} = \frac{1372 \times 3}{27} = \frac{4116}{27}$$ Therefore: $$f\left(\frac{14}{3}\right) = \frac{2744}{27} - \frac{4116}{27} = -\frac{1372}{27}$$ 5. **Final answers:** - Smaller x-value point: $$(0, 0)$$ - Larger x-value point: $$\left(\frac{14}{3}, -\frac{1372}{27}\right)$$