What are the values and types of the critical points, if any, of #f(x) = abs(x^2-1)#?

Question

Cameron Alexander · Accepted Answer

The critical numbers are #-1# and #1# where #f'(x)# does not exist.

Depending on the terminology you are using that might make the critical points #-1# and #1#, or you might have been taught to say they are #(-1,0)# and #(1,0)#. Note that #f(x) >=0# for all #x#. Both #f(-1)# and #f(1)# are #0#, so both are locations of the absolute minimum.

Axel Alvarez · Answer

undefined To find the critical points of $ f(x) = |x^2 - 1| $, we first need to find its derivative and then determine where the derivative is equal to zero or undefined. The derivative of $ |x^2 - 1| $ can be found using the chain rule: $$ f'(x) = \frac{d}{dx} |x^2 - 1| = \frac{d}{dx} \sqrt{(x^2 - 1)^2} $$ $$ f'(x) = \frac{1}{2\sqrt{(x^2 - 1)^2}} \cdot \frac{d}{dx} (x^2 - 1)^2 $$ $$ f'(x) = \frac{2(x^2 - 1) \cdot 2x}{2\sqrt{(x^2 - 1)^2}} $$ $$ f'(x) = \frac{4x(x^2 - 1)}{2|x^2 - 1|} $$ $$ f'(x) = 2x \quad \text{if} \quad x \neq \pm 1 $$ $$ f'(x) = \text{undefined} \quad \text{if} \quad x = \pm 1 $$ The critical points occur where the derivative is equal to zero or undefined: 1. When $ f'(x) = 2x = 0 $, $ x = 0 $. 2. When $ x = \pm 1 $, the derivative is undefined. Now, we need to determine the type of critical point at $ x = 0 $. We can do this by analyzing the behavior of $ f'(x) $ around $ x = 0 $. For this, we can use the first derivative test. Since $ f'(x) = 2x $ is positive for $ x > 0 $ and negative for $ x < 0 $, the function $ f(x) $ is increasing for $ x > 0 $ and decreasing for $ x < 0 $. Therefore, $ x = 0 $ is a local minimum. So, the critical points are: - $ x = 0 $ (local minimum) The corresponding values of the function at these critical points are: - $ f(0) = |0^2 - 1| = 1 $