The function $f(x) = x^2$ is not injective when its domain is the set of all real numbers, but it is injective when its domain is restricted to the set of positive real numbers. We need to explain why this is the case, using examples.

1. Problem Description

The function $f(x) = x^2$ is not injective when its domain is the set of all real numbers, but it is injective when its domain is restricted to the set of positive real numbers. We need to explain why this is the case, using examples.

2. Solution Steps

A function is injective (one-to-one) if for any two distinct values $x_1$ and $x_2$ in its domain, $f(x_1) \ne f(x_2)$. Equivalently, if $f(x_1) = f(x_2)$, then $x_1 = x_2$.

When the domain of $f(x) = x^2$ is the set of all real numbers, we can find distinct values that produce the same output. For example:

$f(2) = 2^2 = 4$

$f(-2) = (-2)^2 = 4$

Since $f(2) = f(-2) = 4$ but $2 \ne -2$, the function is not injective when the domain is the set of all real numbers.

However, when the domain of $f(x) = x^2$ is restricted to the set of positive real numbers, then for any positive real numbers $x_1$ and $x_2$, if $f(x_1) = f(x_2)$, then $x_1^2 = x_2^2$. Taking the square root of both sides gives us $\sqrt{x_1^2} = \sqrt{x_2^2}$. Since $x_1$ and $x_2$ are positive, we have $|x_1| = x_1$ and $|x_2| = x_2$, so $x_1 = x_2$. Therefore, the function is injective when the domain is the set of positive real numbers.

For example, let's say $f(x_1) = f(x_2) = 9$, and $x_1$ and $x_2$ are positive real numbers. Then $x_1^2 = 9$ and $x_2^2 = 9$. Since $x_1$ and $x_2$ are positive, we must have $x_1 = \sqrt{9} = 3$ and $x_2 = \sqrt{9} = 3$. Thus $x_1 = x_2 = 3$.

3. Final Answer

The function $f(x) = x^2$ is not injective when its domain is the set of all real numbers because different values like $2$ and $-2$ can map to the same output ($4$). However, when the domain is restricted to the positive real numbers, the function becomes injective because if $f(x_1) = f(x_2)$, then $x_1^2 = x_2^2$, and since both $x_1$ and $x_2$ are positive, it must be that $x_1 = x_2$.