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Exercise 6

We are given $f : \mathbb{R} \to \mathbb{R}$, where $f(x) = \frac{3}{1+x}$.

1. Calculate $f(\{2, -1, 2\})$:

Substitute $x = 2$, $x = -1$, and $x = 2$ into $f(x)$:
$f(x) = \frac{3}{1+x}$

• For $x = 2$:
  $f(2) = \frac{3}{1+2} = \frac{3}{3} = 1$

• For $x = -1$:
  $f(-1) = \frac{3}{1+(-1)} = \frac{3}{0} \quad \text{(undefined, division by zero).}$

• For $x = 2$ (repetition):
  $f(2) = 1 \quad \text{(as calculated above).}$

Answer: $f(\{2, -1, 2\}) = \{1, \text{undefined}\}$.
The function is undefined at $x = -1$.

2. Is $f$ injective or surjective?

• Injective (One-to-One):
  A function $f$ is injective if $f(x_1) = f(x_2)$ implies $x_1 = x_2$.

For $f(x) = \frac{3}{1+x}$:
$f(x_1) = f(x_2) \implies \frac{3}{1+x_1} = \frac{3}{1+x_2}$ $\implies 1+x_1 = 1+x_2 \quad (\text{cancel the numerator, as } 3 \neq 0)$ $\implies x_1 = x_2.$

Thus, $f$ is injective.

• Surjective (Onto):
  A function $f : \mathbb{R} \to \mathbb{R}$ is surjective if for every $y \in \mathbb{R}$, there exists $x \in \mathbb{R}$ such that $f(x) = y$.

Let $y = f(x) = \frac{3}{1+x}$. Solve for $x$:
$y = \frac{3}{1+x} \implies y(1+x) = 3 \implies y + xy = 3 \implies xy = 3 - y$ $x = \frac{3-y}{y}.$

For $y = 0$, $x$ is undefined (division by zero). Therefore, $f$ is not surjective, as $y = 0$ is not in the range.

Answer: $f$ is injective but not surjective.

3. How to choose the starting and ending sets so that $f$ becomes:

• (a) Injective:
  Since $f$ is already injective, no change is required.

• (b) Surjective:
  To make $f$ surjective, restrict the codomain to $\mathbb{R} \setminus \{0\}$, as $y = 0$ is not achievable by $f(x) = \frac{3}{1+x}$.

• (c) Bijective:
  To make $f$ bijective (both injective and surjective), restrict the codomain to $\mathbb{R} \setminus \{0\}$.

4. If $f$ is invertible, find the inverse of $f$:

To find $f^{-1}$, let $y = f(x) = \frac{3}{1+x}$:
$y = \frac{3}{1+x} \implies y(1+x) = 3 \implies y + xy = 3 \implies xy = 3-y$ $x = \frac{3-y}{y}.$

Thus, the inverse function is: $f^{-1}(y) = \frac{3-y}{y}, \quad y \neq 0.$

Key Notes:

1. The function $f(x) = \frac{3}{1+x}$ is undefined for $x = -1$, as the denominator becomes zero.
2. $f$ is injective but not surjective over $\mathbb{R} \to \mathbb{R}$.
3. The inverse function is valid for $y \neq 0$.

Questions for Further Practice:

1. Prove that $f(x) = \frac{3}{1+x}$ is monotonic over its domain.
2. Verify whether $f$ is bounded (above or below) for $x > -1$.
3. What happens to $f(x)$ as $x \to \infty$ and $x \to -1^+$?
4. Modify $f(x)$ such that it becomes surjective over $\mathbb{R} \to \mathbb{R}$.
5. Analyze the continuity of $f(x)$ at $x = -1$.

Tip: Always check the domain and codomain when determining injectivity, surjectivity, or bijectivity of a function.