Question

Which of the following functions is not invertible?

• A. $f: R \to R, f(x) = 3x + 1$
• B. $f: R \to [0, \infty), f(x) = x^2$
• C. $f: R^+ \to R^+, f(x) = \frac{1}{x^3}$
• D. None of these

Hint:

The invertibility of $x^2$ depends entirely on its domain. If the domain were restricted to $[0, \infty)$, it would be invertible (becoming $\sqrt{x}$). But on the full real number set $R$, the negative values "mirror" the positive ones, breaking injectivity.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
A function is invertible if and only if it is bijective, meaning it is both injective (one-to-one) and surjective (onto). If a function fails the Horizontal Line Test (meaning multiple $x$ values result in the same $y$ value), it is not injective and thus not invertible.

Step 2: Detailed Explanation:
1. Option (a): $f(x) = 3x + 1$ is a linear function. Every $x$ gives a unique $y$, and every $y \in R$ has a pre-image. It is bijective and invertible. 2. Option (b): $f(x) = x^2$ with domain $R$. Check injectivity: - $f(2) = 4$ and $f(-2) = 4$. - Since $f(2) = f(-2)$ but $2 \neq -2$, the function is many-to-one. - Because it is not injective, it cannot be bijective, making it not invertible. 3. Option (c): $f(x) = 1/x^3$ with domain $R^+$. In this domain, $x$ is always positive. The function is strictly decreasing and one-to-one. It is invertible.

Step 3: Final Answer
The function $f(x) = x^2$ defined on $R$ is not invertible.