Question

let f(x) = square root (x^2-2x) a. find f'(x) b.find critical numbers (points) of f

Answer 1

a) f'(x)=1/2(x^2-2x)^(-1/2) (2x-2) = (x-1)(x^2-2x)^(-1/2)

Answer 2

b) 0=(x-1)(x^2-2x)^(-1/2) = (x-1)/square root of (x^2-2x) and then, using algebra,
0=x-1 so x=1
therefore, 1 is the critical number

Answer 3

thanks