How do you find the critical numbers of #y = abs(x^2 -1)#?

Question

Charles Anctil · Accepted Answer

The critical numbers are #-1, 0, 1#

A critical number for #f# is a number #c#, in the domain of #f# at which #f'(c)# does not exist or #f'(c)=0#.

Therefore, we begin by finding #f'(x)# for #f(x) = abs(x^2-1)#.

NOte that the domain of #f# is #(-oo,oo)#.

#f(x) = abs(x^2-1) = {(x^2-1,"if",x^2-1 >= 0),(-(x^2-1),"if",x^2-1 < 0) :}#.

Investigating the sign of #x^2-1# shows that it is positive if #x < 1# or #x > 1# and it is negative if #-1 < x < 1#.

Therefore,

#f(x) = {(x^2-1,"if",x <= -1),(-x^2+1,"if",-1 < x < 1),(x^2-1,"if",x >= 1) :}#.

Differentiating each piece gets us

#f'(x) = {(2x,"if",x < -1),(-2x,"if",-1 < x < 1),(2x,"if",x > 1) :}#.

At the 'joints' of #x=-1# and #x=1# the left and right derivatives do not agree, so the derivative does not exist.

#-1# and #1# are critical numbers for #f#.

Furthemore, #f'(x) = 0# at #x=0#, so #0# is also a critical number for #f#.

The critical numbers are #-1, 0, 1#

Emily Ammerman · Answer

undefined To find the critical numbers of $ y = |x^2 - 1| $, we first need to determine where the derivative is either zero or undefined. Then, we check those values to see if they correspond to local extrema or points where the function changes direction.

The derivative of $ |x^2 - 1| $ can be found by using the Chain Rule and the fact that the derivative of the absolute value function |u| is $ \frac{{du}}{{dx}} $ if $ u $ is differentiable and $ \frac{{-du}}{{dx}} $ if $ u $ is not differentiable.

So, differentiate $ |x^2 - 1| $ using the Chain Rule:

$$ \frac{{dy}}{{dx}} = \frac{{d}}{{dx}}|x^2 - 1| = \frac{{d}}{{dx}}(x^2 - 1) $$ for $ x^2 - 1 > 0 $,

$$ \frac{{dy}}{{dx}} = \frac{{d}}{{dx}}-(x^2 - 1) $$ for $ x^2 - 1 < 0 $.

Solve these two derivatives separately to find critical points.