Answer:
3rd option: {0}
Step-by-step explanation:
Critical points of a function are where [tex]f'(x)=0[/tex], therefore:
[tex]f(x)=\frac{x^2-1}{x^2-4}\\\\f'(x)=\frac{(x^2-4)(2x)-(x^2-1)(2x)}{(x^2-4)^2}\\ \\f'(x)=-\frac{6x}{(x^2-4)^2}\\\\0=-\frac{6x}{(x^2-4)^2}\\\\0=-6x\\\\x=0[/tex]
This means that the set of x-values that identifies all critical points of the function is {0}.
