To find the relative extreme points of the function $G(x) = x^3 - 2x^2 - 4x - 2$, we need to follow these steps:

Step 1: Find the first derivative

The first derivative, $G'(x)$, gives us the critical points where the slope is zero (possible relative extrema). $G'(x) = \frac{d}{dx}(x^3 - 2x^2 - 4x - 2) = 3x^2 - 4x - 4$

Step 2: Set the first derivative equal to zero to find critical points

Now solve $G'(x) = 0$: $3x^2 - 4x - 4 = 0$ This is a quadratic equation, and we can use the quadratic formula to solve for $x$: $x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(3)(-4)}}{2(3)}$ $x = \frac{4 \pm \sqrt{16 + 48}}{6}$ $x = \frac{4 \pm \sqrt{64}}{6}$ $x = \frac{4 \pm 8}{6}$ Thus, we get two solutions: $x = \frac{4 + 8}{6} = 2 \quad \text{and} \quad x = \frac{4 - 8}{6} = -\frac{2}{3}$

Step 3: Determine whether these critical points are maxima, minima, or neither

We use the second derivative test to determine the nature of the critical points. The second derivative of $G(x)$ is: $G''(x) = \frac{d}{dx}(3x^2 - 4x - 4) = 6x - 4$

• For $x = 2$: $G''(2) = 6(2) - 4 = 12 - 4 = 8 \quad (\text{positive, so it's a relative minimum at } x = 2)$

• For $x = -\frac{2}{3}$: $G''\left(-\frac{2}{3}\right) = 6\left(-\frac{2}{3}\right) - 4 = -4 - 4 = -8 \quad (\text{negative, so it's a relative maximum at } x = -\frac{2}{3})$

Step 4: Find the y-coordinates of the relative extrema

• For $x = 2$: $G(2) = (2)^3 - 2(2)^2 - 4(2) - 2 = 8 - 8 - 8 - 2 = -10$ So, the relative minimum is $(2, -10)$.

• For $x = -\frac{2}{3}$: $G\left(-\frac{2}{3}\right) = \left(-\frac{2}{3}\right)^3 - 2\left(-\frac{2}{3}\right)^2 - 4\left(-\frac{2}{3}\right) - 2$ $= -\frac{8}{27} - 2\left(\frac{4}{9}\right) + \frac{8}{3} - 2$ Simplifying this expression results in $G\left(-\frac{2}{3}\right) \approx 1.037$.

Thus, the relative maximum is approximately $\left(-\frac{2}{3}, 1.037\right)$.

Conclusion

• The relative minimum point is $(2, -10)$.
• The relative maximum point is approximately $\left(-\frac{2}{3}, 1.037\right)$.

Graph

The graph of the function $G(x) = x^3 - 2x^2 - 4x - 2$ has a relative maximum near $x = -\frac{2}{3}$ and a relative minimum at $x = 2$. The function behaves like a cubic function, increasing and decreasing between these critical points.

Would you like me to generate a sketch of the graph?