Question

Determine whether the function \(f : \mathbb{R} \to \mathbb{R}\) defined by \(f(x) = 2x^3 - 7\), \(\forall\, x \in \mathbb{R}\) (Real numbers) is bijective or not.

Hint:

Show \(f(x_1) = f(x_2)\) gives \(x_1 = x_2\), and solve \(y = 2x^3 - 7\) for \(x\) to show every real \(y\) is reached.

Answer 1

Step 1: Plan:
A function is bijective when it is both one-one (injective) and onto (surjective). We check each.

Step 2: Check one-one:
Take \(x_1, x_2 \in \mathbb{R}\) with \(f(x_1) = f(x_2)\).
\[ 2x_1^3 - 7 = 2x_2^3 - 7 \Rightarrow x_1^3 = x_2^3 \]
Every real number has exactly one real cube root, so \(x_1 = x_2\).
So f is one-one.

Step 3: Check onto:
Let \(y \in \mathbb{R}\) be any element of the codomain. Put \(y = 2x^3 - 7\).
\[ x^3 = \frac{y + 7}{2} \Rightarrow x = \left(\frac{y + 7}{2}\right)^{1/3} \]
This \(x\) is a real number for every real \(y\).
Check: \(f(x) = 2 \cdot \frac{y+7}{2} - 7 = y\). So every \(y\) has a pre-image.
So f is onto.

Final Answer:
f is both one-one and onto, so f is bijective.