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

1 Answer
May 20, 2017


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


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.