Find the critical numbers of f, if any, find the open intervals on which the function is increasing/decreasing, apply the First Derivative Test to identify all relative extrema for g(x) = x2 − 2x − 8.

$$g(x) = x^2 - 2x - 8$$ $$g'(x) = \frac{d}{dx}(x^2 - 2x - 8) = 2x - 2$$

$$2x - 2 = 0$$ $$2x = 2$$ $$x = 1$$

The critical number is $x=1$.

We test values in the intervals $(-\infty, 1)$ and $(1, \infty)$.

• For $x < 1$, let's pick $x=0$: $g'(0) = 2(0) - 2 = -2$. Since $g'(0) < 0$, $g(x)$ is decreasing on $(-\infty, 1)$.
• For $x > 1$, let's pick $x=2$: $g'(2) = 2(2) - 2 = 4 - 2 = 2$. Since $g'(2) > 0$, $g(x)$ is increasing on $(1, \infty)$.

Since $g'(x)$ changes from negative to positive at $x=1$, there is a relative minimum at $x=1$.
The value of the function at $x=1$ is:

$$g(1) = (1)^2 - 2(1) - 8 = 1 - 2 - 8 = -9$$

Summary:

• Critical number: $\boxed{x=1}$
• Increasing interval: $\boxed{(1, \infty)}$
• Decreasing interval: $\boxed{(-\infty, 1)}$
• Relative extrema: $\boxed{\text{Relative minimum at } (1, -9)}$