Question

f(x)=sqrt(x^2-1)/(x-abs(x)) show that for all x on (negative infinity, -1], f(x)>-1/2 and conclude that range of f equals(-1/2, 0] by using the intermediate value theorem

Answer 1

It may help to note that for \(x\in(-\infty,-1],\;we\;have\;|x| = -x\)