Question

If \( f: \mathbb{Z} \to \mathbb{Z} \), defined by \( f(x) = x^3 + 2 \), then the function \( f \) is __________.

Hint:

To check injectivity, solve \( f(a) = f(b) \) and see if it leads to \( a = b \). If true, the function is one-to-one.

Answer 1

Step 1: Understanding one-to-one (injective) functions. A function \( f \) is one-to-one (injective) if: \[ f(a) = f(b) \Rightarrow a = b. \] 

Step 2: Checking injectivity of \( f(x) \). Given: \[ f(x) = x^3 + 2. \] Let \( f(a) = f(b) \): \[ a^3 + 2 = b^3 + 2. \] Canceling \( 2 \) from both sides: \[ a^3 = b^3. \] Taking the cube root: \[ a = b. \] Since the function satisfies the condition for injectivity, it is one-to-one. 

Step 3: Checking onto (surjective) property. For surjectivity, \( f(x) \) must cover all integers. Since cube functions do not necessarily map to all integers, \( f(x) \) is not necessarily onto over \( \mathbb{Z} \). 

Conclusion: The function \( f(x) \) is one-to-one.