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 f(x) = x2 − 8x.

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

$$2x - 8 = 0$$ $$2x = 8$$ $$x = 4$$

The critical number is $x=4$.

We test values in the intervals $(-\infty, 4)$ and $(4, \infty)$.

• For $x < 4$, let's pick $x=0$: $f'(0) = 2(0) - 8 = -8$. Since $f'(0) < 0$, $f(x)$ is decreasing on $(-\infty, 4)$.
• For $x > 4$, let's pick $x=5$: $f'(5) = 2(5) - 8 = 10 - 8 = 2$. Since $f'(5) > 0$, $f(x)$ is increasing on $(4, \infty)$.

Since $f'(x)$ changes from negative to positive at $x=4$, there is a relative minimum at $x=4$.
The value of the function at $x=4$ is:

$$f(4) = (4)^2 - 8(4) = 16 - 32 = -16$$

Summary:

• Critical number: $\boxed{x=4}$
• Increasing interval: $\boxed{(4, \infty)}$
• Decreasing interval: $\boxed{(-\infty, 4)}$
• Relative extrema: $\boxed{\text{Relative minimum at } (4, -16)}$