Key Concepts and Formulas

• Injective (One-to-one) Function: A function $f: A \to B$ is injective if for any $x_1, x_2 \in A$, $f(x_1) = f(x_2)$ implies $x_1 = x_2$. This means distinct elements in the domain map to distinct elements in the codomain.
• Surjective (Onto) Function: A function $f: A \to B$ is surjective if for every element $y \in B$, there exists at least one element $x \in A$ such that $f(x) = y$. This means every element in the codomain is mapped to by at least one element in the domain.
• Absolute Value Function: $|a| = a$ if $a \ge 0$, and $|a| = -a$ if $a < 0$.

Step-by-Step Solution

Step 1: Understand the Domain and Codomain of the Function The function is given by $f(x) = \left| {1 - {1 \over x}} \right|$. The domain is $f : (0, \infty) \to (0, \infty)$. This means the input $x$ can be any positive real number, and the output $f(x)$ must also be a positive real number.

Step 2: Analyze the Injectivity of the Function To check for injectivity, we need to see if $f(x_1) = f(x_2)$ implies $x_1 = x_2$ for $x_1, x_2 \in (0, \infty)$.

Let's set $f(x_1) = f(x_2)$: $\left| {1 - {1 \over x_1}} \right| = \left| {1 - {1 \over x_2}} \right|$

This equation implies two possibilities: Case 1: $1 - {1 \over x_1} = 1 - {1 \over x_2}$ Subtracting 1 from both sides gives: $-{1 \over x_1} = -{1 \over x_2}$ Multiplying by $-1$: ${1 \over x_1} = {1 \over x_2}$ Taking the reciprocal of both sides: $x_1 = x_2$ This case supports injectivity.

Case 2: $1 - {1 \over x_1} = - (1 - {1 \over x_2})$ $1 - {1 \over x_1} = -1 + {1 \over x_2}$ Rearranging the terms to group $x$ terms and constants: $1 + 1 = {1 \over x_1} + {1 \over x_2}$ $2 = {x_2 + x_1 \over x_1 x_2}$ $2 x_1 x_2 = x_1 + x_2$

Let's test if we can find distinct $x_1$ and $x_2$ in the domain $(0, \infty)$ that satisfy this equation and also $f(x_1) = f(x_2)$. Consider $x_1 = 1$. Then $f(1) = |1 - 1/1| = |0| = 0$. However, the codomain is $(0, \infty)$, so $f(1)=0$ is not allowed. This suggests we need to be careful.

Let's re-examine the function $f(x) = \left| {1 - {1 \over x}} \right|$. For $x \in (0, \infty)$: If $x=1$, $f(1) = |1-1| = 0$. But the codomain is $(0, \infty)$. This means $x=1$ is not in the domain that maps to the codomain $(0, \infty)$. Let's reconsider the domain of $f$. The problem states $f : (0, \infty) \to (0, \infty)$. This means that for all $x \in (0, \infty)$, $f(x)$ must be in $(0, \infty)$. So, $\left| {1 - {1 \over x}} \right| > 0$. This implies $1 - {1 \over x} \ne 0$, so ${1 \over x} \ne 1$, which means $x \ne 1$. Therefore, the effective domain for which the function maps into $(0, \infty)$ is $(0, 1) \cup (1, \infty)$.

Let's check injectivity again with the effective domain $(0, 1) \cup (1, \infty)$. We are looking for $x_1, x_2 \in (0, 1) \cup (1, \infty)$ such that $f(x_1) = f(x_2)$ and $x_1 \ne x_2$. From Case 2: $2 x_1 x_2 = x_1 + x_2$. Divide by $x_1 x_2$ (since $x_1, x_2 > 0$, $x_1 x_2 \ne 0$): $2 = {x_1 \over x_1 x_2} + {x_2 \over x_1 x_2}$ $2 = {1 \over x_2} + {1 \over x_1}$ $2 = {1 \over x_1} + {1 \over x_2}$

Let's try to find $x_1 \ne x_2$ in $(0, 1) \cup (1, \infty)$ satisfying this. If we choose $x_1 = 2$, then ${1 \over x_1} = {1 \over 2}$. $2 = {1 \over 2} + {1 \over x_2}$ ${1 \over x_2} = 2 - {1 \over 2} = {3 \over 2}$ $x_2 = {2 \over 3}$ Both $x_1 = 2$ and $x_2 = 2/3$ are in the domain $(0, 1) \cup (1, \infty)$. Let's check the function values: $f(2) = |1 - 1/2| = |1/2| = 1/2$. $f(2/3) = |1 - 1/(2/3)| = |1 - 3/2| = |-1/2| = 1/2$. Since $f(2) = f(2/3)$ but $2 \ne 2/3$, the function is not injective.

Step 3: Analyze the Surjectivity of the Function To check for surjectivity, we need to see if for every $y \in (0, \infty)$, there exists an $x \in (0, \infty)$ such that $f(x) = y$. We are given $f(x) = \left| {1 - {1 \over x}} \right|$ and the codomain is $(0, \infty)$. So, we need to solve $y = \left| {1 - {1 \over x}} \right|$ for $x \in (0, \infty)$, for any $y \in (0, \infty)$.

This equation is equivalent to: $1 - {1 \over x} = y$ or $1 - {1 \over x} = -y$

Case A: $1 - {1 \over x} = y$ $1 - y = {1 \over x}$ $x = {1 \over {1 - y}}$ For $x$ to be in the domain $(0, \infty)$, we need $x > 0$. Since $y \in (0, \infty)$, $y > 0$. If $y > 1$, then $1-y < 0$, so $x = {1 \over {1 - y}} < 0$. This value of $x$ is not in the domain. If $0 < y < 1$, then $1-y > 0$, so $x = {1 \over {1 - y}} > 0$. This value of $x$ is in the domain. For example, if $y = 1/2$, then $x = {1 \over {1 - 1/2}} = {1 \over {1/2}} = 2$. And $f(2) = |1 - 1/2| = 1/2$.

Case B: $1 - {1 \over x} = -y$ $1 + y = {1 \over x}$ $x = {1 \over {1 + y}}$ For $x$ to be in the domain $(0, \infty)$, we need $x > 0$. Since $y \in (0, \infty)$, $y > 0$. Therefore, $1+y > 1$, which means $x = {1 \over {1 + y}}$ is always between 0 and 1 ($0 < x < 1$). This value of $x$ is always in the domain $(0, \infty)$. For example, if $y = 1/2$, then $x = {1 \over {1 + 1/2}} = {1 \over {3/2}} = 2/3$. And $f(2/3) = |1 - 3/2| = |-1/2| = 1/2$.

Now, let's consider the range of $f(x) = \left| {1 - {1 \over x}} \right|$ for $x \in (0, \infty)$. If $x \in (0, 1)$, then $1/x > 1$, so $1 - 1/x < 0$. $f(x) = -(1 - 1/x) = 1/x - 1$. As $x \to 0^+$, $1/x \to \infty$, so $f(x) \to \infty$. As $x \to 1^-$, $1/x \to 1^+$, so $1/x - 1 \to 0^+$. So, for $x \in (0, 1)$, the range of $f(x)$ is $(0, \infty)$.

If $x = 1$, $f(1) = |1 - 1| = 0$. But the codomain is $(0, \infty)$, so $f(1)$ is not in the codomain.

If $x \in (1, \infty)$, then $0 < 1/x < 1$, so $0 < 1 - 1/x < 1$. $f(x) = 1 - 1/x$. As $x \to 1^+$, $1/x \to 1^-$, so $1 - 1/x \to 0^+$. As $x \to \infty$, $1/x \to 0^+$, so $1 - 1/x \to 1^-$. So, for $x \in (1, \infty)$, the range of $f(x)$ is $(0, 1)$.

Combining the ranges: For $x \in (0, 1)$, the range is $(0, \infty)$. For $x \in (1, \infty)$, the range is $(0, 1)$. The overall range of $f(x)$ for $x \in (0, 1) \cup (1, \infty)$ is $(0, \infty)$.

Now consider the codomain, which is $(0, \infty)$. For any $y \in (0, \infty)$, we need to find an $x \in (0, \infty)$ such that $f(x) = y$.

If $y \in (0, 1)$, we can use $x = {1 \over {1 + y}}$ (from Case B), which is in $(0, 1/2)$, and thus in $(0, \infty)$. Or we can use $x = {1 \over {1 - y}}$ (from Case A), which is in $(1, \infty)$, and thus in $(0, \infty)$. For example, if $y = 1/2$, we found $x=2$ and $x=2/3$. Both are in $(0, \infty)$.

If $y \in [1, \infty)$, we cannot use Case A ($x = {1 \over {1 - y}}$) because $1-y \le 0$, leading to $x \le 0$ or $x$ undefined. However, we can use Case B ($x = {1 \over {1 + y}}$). Since $y \ge 1$, $1+y \ge 2$, so $0 < x = {1 \over {1 + y}} \le 1/2$. This $x$ is in $(0, \infty)$. For example, if $y = 2$, then $x = {1 \over {1 + 2}} = {1 \over 3}$. Let's check $f(1/3) = |1 - 1/(1/3)| = |1 - 3| = |-2| = 2$. So, for $y=2$, we found an $x=1/3$ in the domain $(0, \infty)$.

Thus, for every $y \in (0, \infty)$, there exists an $x \in (0, \infty)$ such that $f(x) = y$. Therefore, the function is surjective.

Step 4: Conclude the Nature of the Function From Step 2, we found that the function is not injective because $f(2) = f(2/3) = 1/2$ with $2 \ne 2/3$. From Step 3, we found that the function is surjective because for every $y \in (0, \infty)$, there exists an $x \in (0, \infty)$ such that $f(x) = y$.

Therefore, the function is not injective but it is surjective.

Common Mistakes & Tips

• Ignoring the Codomain: Always ensure that the output of the function for any input in the domain falls within the specified codomain. If not, the function might not be well-defined for that domain-codomain pair, or it might affect surjectivity.
• Confusing Domain and Codomain: The question specifies $f : (0, \infty) \to (0, \infty)$. This means inputs are positive, and outputs must be positive. The value $f(x)=0$ is not in the codomain.
• Absolute Value Properties: Remember that $|a|=|b|$ implies $a=b$ or $a=-b$. This is crucial for solving equations involving absolute values.
• Graphical Interpretation: For injectivity, a horizontal line should intersect the graph at most once. For surjectivity, a horizontal line at any $y$ value in the codomain should intersect the graph at least once.

Summary We analyzed the function $f(x) = \left| {1 - {1 \over x}} \right|$ with domain and codomain $(0, \infty)$. We determined that the function is not injective because different inputs can produce the same output (e.g., $f(2) = f(2/3)$). We also determined that the function is surjective because every value in the codomain $(0, \infty)$ can be achieved as an output for some input in the domain $(0, \infty)$. Therefore, the function is neither injective nor surjective.

The final answer is \boxed{A}.