Question

Show that the function \( f : \mathbb{R} \to \mathbb{R} \) defined by \( f(x) = 4x^3 - 5 \), \( \forall x \in \mathbb{R} \), is one-one and onto.

Hint:

Quick Tip: To prove a function is one-one, assume \( f(x_1) = f(x_2) \) and show that it leads to \( x_1 = x_2 \). To prove a function is onto, solve for \( x \) in terms of \( y \) to show every value of \( y \) corresponds to some \( x \).

Answer 1

Demonstration of one-to-one property:
A function is defined as one-to-one (or injective) if, for any \( x_1, x_2 \in \mathbb{R} \), the condition \( f(x_1) = f(x_2) \) implies \( x_1 = x_2 \).
Assume \( f(x_1) = f(x_2) \). This leads to:
\[ 4x_1^3 - 5 = 4x_2^3 - 5 \] Upon simplification:
\[ 4x_1^3 = 4x_2^3 \] \[ x_1^3 = x_2^3 \] Taking the cube root of both sides yields:
\[ x_1 = x_2 \] Consequently, the function \( f(x) = 4x^3 - 5 \) is demonstrated to be one-to-one (injective).
Demonstration of onto property:
A function is considered onto (or surjective) if for every element \( y \in \mathbb{R} \), there exists an element \( x \in \mathbb{R} \) such that \( f(x) = y \).
Let \( y \in \mathbb{R} \) be an arbitrary value. We aim to find an \( x \in \mathbb{R} \) satisfying:
\[ f(x) = y \] Substituting the function definition:
\[ 4x^3 - 5 = y \] Solving for \( x \):
\[ 4x^3 = y + 5 \] \[ x^3 = \frac{y + 5}{4} \] \[ x = \sqrt[3]{\frac{y + 5}{4}} \] This shows that for any \( y \in \mathbb{R} \), a corresponding \( x \in \mathbb{R} \) exists such that \( f(x) = y \). Therefore, the function is onto (surjective).
Conclusion: As the function exhibits both one-to-one and onto properties, it is classified as bijective.