Question

How do you find the critical numbers of an absolute value equation f(x) = |x + 3| - 1?

Answer 1

There is one critical point: `x = -3, y = -1`

Explanation:

Critical points are minima, maxima, and points where `f'(x) = 0`.

Since this is a v-shaped absolute value function, there is no point where f'(x) = 0. However, there is one extremum. It is a minimum located at `x = -3`, and it is the point `(-3, -1)`.

You can figure this out by looking at how `f(x) = |x+3|-1` is a modification of its parent function `f(x) = |x|`. It is shifted left 3 and down 1, so the local minimum from the parent function is also shifted left 3 and down 1.