mycroft holmes
Last Activity: 15 Years ago
thats a shockingly atrocious reply by the expert
Its obvious that g(x) is non-negative. Hence for negative real values there is no x such that g(x) = y. So onto is ruled out.
Again if x and y are two real numbers such that f(x) = - f(y) we have g(x) = g(y) though x /= y . So one-one is out.
Continuous is right, as if a sequence {x} is convergent to l, {|x|} is convergent to |l|
Differentiability-the critical points are where f(x) = 0. If f(x) changes sign in an interval containing this point and if f'(x) /= 0 at this point then we have a point of non-differentiability.