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

To prove that the function is one-one:
A function is said to be one-one (or injective) if for all \( x_1, x_2 \in \mathbb{R} \), whenever \( f(x_1) = f(x_2) \), we have \( x_1 = x_2 \).
Let us assume that \( f(x_1) = f(x_2) \). Then,
\[ 4x_1^3 - 5 = 4x_2^3 - 5 \] Simplifying this, we get: \[ 4x_1^3 = 4x_2^3 \] \[ x_1^3 = x_2^3 \] Taking the cube root on both sides: \[ x_1 = x_2 \] Thus, the function \( f(x) = 4x^3 - 5 \) is one-one (injective).
To prove that the function is onto:
A function is said to be onto (or surjective) if for every \( y \in \mathbb{R} \), there exists an \( x \in \mathbb{R} \) such that \( f(x) = y \).
Let \( y \in \mathbb{R} \) be arbitrary. We need to find an \( x \in \mathbb{R} \) such that: \[ f(x) = y \] Substitute the expression for \( f(x) \): \[ 4x^3 - 5 = y \] Solving for \( x \): \[ 4x^3 = y + 5 \] \[ x^3 = \frac{y + 5}{4} \] \[ x = \sqrt[3]{\frac{y + 5}{4}} \] Thus, for every \( y \in \mathbb{R} \), there exists an \( x \in \mathbb{R} \) such that \( f(x) = y \). Therefore, the function is onto (surjective).
Conclusion: Since the function is both one-one and onto, it is bijective.