Question

Let $f, g : \mathbb{R} \to \mathbb{R}$ be functions. If $g$ is continuous, then which one of the following cases implies that $f$ is continuous?

• A. $g(x) = (f(x))^3$
• B. $g(x) = |f(x)|$
• C. $g(x) = (f(x))^2$
• D. $g(x) = \sin(f(x))$

Hint:

An odd power function like \(t^3\) is a homeomorphism of \(\mathbb{R}\)., meaning both the function and its inverse are continuous.
This ensures that taking the cube root preserves continuity.
Even powers or periodic functions lose information about the sign or interval, allowing discontinuities to be hidden.

Answer 1

The Correct Option is A

To determine which of the given options implies that the function \(f\) is continuous when \(g\) is continuous, we need to analyze each case individually. We are given that \(g\) is a continuous function, and we want to deduce under which transformation this would imply the continuity of \(f\). Let's consider each case:

1. \(g(x) = (f(x))^3\):
   If \(g(x) = (f(x))^3\) is continuous, then the cube root transformation which is continuous and invertible implies \(f(x)\) is continuous. This is because the cube root function is continuous across all real numbers, and the only way for the composition to remain continuous is if \(f(x)\) itself is continuous.
2. \(g(x) = |f(x)|\):
   \(|f(x)|\) is continuous whenever \(f(x)\) is continuous. However, the converse does not necessarily imply that \(f(x)\) is continuous. Consider the function \(f(x) =\)-1 for \(x \leq 0\) and 1 for \(x > 0\). Here, \(|f(x)| = 1\) is continuous, even though \(f(x)\) is not continuous.
3. \(g(x) = (f(x))^2\):
   Similar to the absolute value case, \((f(x))^2\) being continuous does not imply \(f(x)\) is continuous. The squaring function does not differentiate between positive and negative values, allowing a discontinuous \(f(x)\) to result in a continuous \(g(x)\). The same counterexample as above applies, where \(g(x) = 1\) is continuous.
4. \(g(x) = \sin(f(x))\):
   The sine function is continuous for all \(x\) in \(\mathbb{R}\) and periodic, thus a discontinuity in \(f(x)\) could still result in a continuous composition \(g(x)\). Similar examples can be devised where \(f(x)\) is not continuous, but since sine is periodic, \(g(x)\) is.

Based on this analysis, the correct answer is \(g(x) = (f(x))^3\). In this case, the continuity of \(g(x)\) directly implies that \(f(x)\) must be continuous.