If f(x) is differentiable for all x and the range of f(x) is (-oo, oo), then if g(x) = |f(x)|, g(x) must have_

Question

an absolute minimum and the absolute minimum will be g(x) = 0


True or False ???

Answer 1

True.
You should probably look this up in your book.
I'm thinking that "since f(x) is differentiable for all x, then it is continuous for all x."
The part in quotes is what you should check on.
Assuming that is true, then the smallest g(x) could be is 0, since the absolute value function is never negative.

Answer 2

This should be true. If the function is differentiable for all x then it must be continuous for all x. With a range of -infinitity to positive infinity, it would have to cross the x axis. At such points the absolute value of the function would be zero and this would be an absolute minimum. Something like f(x)=4 doesn't provide a counterexample because it does not have a range from -infinity to positive infinity. Technically this could be a trick question in its wording though. The absolute minimum value would be g(x)=0, but the absolute minimum would be the point or points where such a vaulue occurs.

Answer 3

True. (Even if f is non-differentiable or non-continuous!)

If the range of f is (-oo, +oo) it certainly contains zero, so there exists a point x for which f(x) = 0. This proves that the absolute minimum of g must be <= 0.

Because by definition g(x) = |f(x)| >= 0, the minimum has to be zero.

Answer 4

False. Counterexample f(x) = 4 for all x. Then g(x) = abs(f(x)) = 4. The absolute minimum of g is not zero.