Solution to Let f:R\to→ R be f(x)= x/1+|x| a) Is f everywhere defined? If not, give the … - Sikademy
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Archangel Macsika

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) = \frac{x}{1+|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) = \frac{x}{1+|x|}

This is a positive number for such x and is a proper fraction, being bounded above by 1.

Observe that f(-x) = \frac{-x}{1+|x|}

We see that f is an odd function whose graph is symmetric about the origin. Therefore…

For x < 0, f(x) = \frac{x}{1-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

\frac{x}{1+x} = \frac{1+x-1}{1+x}

\frac{1+x}{1+x} - \frac{1}{1+x}

= 1 - \frac{1}{1+x}

Assume f(a) = f(b), where a and b are positive.

1 - \frac{1}{1+a} = 1 - \frac{1}{1+b}

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.

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