To determine which function is not invertible, we need to understand the concept of invertibility of a function. A function is invertible if and only if it is a bijective function, meaning it should be both one-to-one (injective) and onto (surjective).
- Option 1: \(f: \mathbb{R} \to \mathbb{R}, \, f(x) = 3x + 1\)
This is a linear function with a non-zero slope, therefore it is one-to-one. For every \(y\) in \(\mathbb{R}\), we can find an \(x\) such that \(f(x) = y\), therefore it is also onto. Hence, this function is invertible. - Option 2: \(f: \mathbb{R} \to [0, \infty), \, f(x) = x^2\)
This function maps every real number to its square. It is not one-to-one because \(f(x) = f(-x)\) for any \(x \neq 0\) (e.g., \(f(2)=f(-2)=4\)). Additionally, it does not map to any negative numbers in its co-domain. Its lack of injectivity makes it not invertible. - Option 3: \(f: \mathbb{R}^+ \to \mathbb{R}^+, \, f(x) = \frac{1}{x^3}\)
This function is one-to-one because different inputs give different outputs. It's also onto because for any positive \(y\), we can find \(x\) such that \(f(x) = y\). Thus, this function is invertible.
The function that is not invertible is Option 2, \(f: \mathbb{R} \to [0, \infty), \, f(x) = x^2\), as it is not one-to-one.