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

1 Answer
John D.
May 20, 2017

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

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