Let f:R\to→ R be f(x)= x/1+|x| a) Is f everywhere defined? If not, give the domain. b) Is f onto? If not give the range. c) Is f one-to-one? Explain. d) Is f invertible? If so, what is f-1?
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Let f(x) =
The domain of f consists of all values of x for which the denominator is nonzero.
Since |x| ≥ 0 for all x, the denominator is at least 1. Therefore, the denominator is never zero. The domain consists of all real numbers.
Let’s try to do this in a way that steers you toward the steps rather than simply doing them for you.
(b) For x > 0, f(x) =
This is a positive number for such x and is a proper fraction, being bounded above by 1.
Observe that f(-x) =
We see that f is an odd function whose graph is symmetric about the origin. Therefore…
For x < 0, f(x) =
This is a negative number for such x and is bounded below by -1.
The range of f is the open interval (-1, 1).
(c) Yes, it is one-to-one. Using the symmetry, we only need to examine the interval where x is positive.
First, observe that
= 1 -
Assume f(a) = f(b), where a and b are positive.
1 - = 1 -
You will see that when you unravel both sides, you wind up with a = b. By the definition of one-to-one, f is one-to-one for x greater than or equal to zero, and by symmetry, f is one-to-one for all real numbers.
(d) Since f is one-to-one, it is invertible.
Again, let x > 0. Solve this for x. I will let that instruction be enough to get you started. For x < 0, you may use the other definition (given above), and solve for x, or you may use symmetry again.