What are the critical points of #(x^2/(x^2-1))#?

Question

William Abdalla · Accepted Answer

I will put f(x) in front of  #(x^2/(x^2-1))# To find the critical number, you must get the first derivative of
f(x) = #(x^2/(x^2-1))#   The first derivative is  #f^'(x) = (-2x)/(x^2-1)^2#  Now you must set  #f^'(x) = 0#, and you must  also find where the #f^'(x)# does not exist (dne). #f^'(x) = 0#                  -------------------------     #f^'(x) #  dne           #(-2x)/(x^2-1)^2# = 0   ------------------#f^'(x)# dne at x = 1 and x = -1 -2x = 0
 x = 0
 --------------------------- now you have to plug them back into the original equation. x = 1 and x = -1 are not critical numbers because they are undefined. x = 0 is a crucial number since it is defined in the original equation.

Isaiah Allen · Answer

undefined The critical points of the function $ \frac{x^2}{x^2 - 1} $ occur where the derivative is either zero or undefined. To find these points, we first find the derivative of the function:

$ \frac{d}{dx} \left( \frac{x^2}{x^2 - 1} \right) = \frac{(x^2 - 1)(2x) - x^2(2x)}{(x^2 - 1)^2} $.

Simplifying this derivative, we get:

$ \frac{d}{dx} \left( \frac{x^2}{x^2 - 1} \right) = \frac{-2x}{(x^2 - 1)^2} $.

Setting the derivative equal to zero and solving for $ x $, we get:

$ -2x = 0 $, which gives us $ x = 0 $.

Therefore, the only critical point of the function $ \frac{x^2}{x^2 - 1} $ is at $ x = 0 $.