Step 1: Understanding the Concept:
We need to distinguish between continuity and uniform continuity on the entire real line \(\mathbb{R}\).
- Continuity at a point \(c\) means for any \(\epsilon > 0\), there exists a \(\delta > 0\) such that if \(|x-c| < \delta\), then \(|f(x)-f(c)| < \epsilon\). Here, \(\delta\) can depend on both \(\epsilon\) and the point \(c\).
- Uniform Continuity on a set \(S\) means for any \(\epsilon > 0\), there exists a \(\delta > 0\) such that for any two points \(x, y \in S\), if \(|x-y| < \delta\), then \(|f(x)-f(y)| < \epsilon\). Here, \(\delta\) depends only on \(\epsilon\), not on the specific location of the points. A function is not uniformly continuous if its slope becomes arbitrarily steep.
Step 3: Detailed Explanation:
Let's analyze each function on \(\mathbb{R}\):
- (A) \(f(x) = \sin x\): This function is continuous everywhere. The derivative is \(f'(x) = \cos x\), which is bounded: \(|\cos x| \le 1\). A function with a bounded derivative on an interval is uniformly continuous on that interval. Since the derivative is bounded on all of \(\mathbb{R}\), \(\sin x\) is uniformly continuous on \(\mathbb{R}\).
- (B) \(f(x) = x^2\): This function is a polynomial, so it is continuous everywhere on \(\mathbb{R}\). To check for uniform continuity, let's look at its derivative, \(f'(x) = 2x\). The derivative is not bounded on \(\mathbb{R}\) (it goes to \(\infty\) as \(x \to \infty\)). This suggests the function is not uniformly continuous.
To prove it formally, let \(\epsilon = 1\). For any \(\delta > 0\), we need to find \(x, y\) with \(|x-y| < \delta\) such that \(|f(x)-f(y)| \ge 1\). Let \(y = x + \delta/2\). Then \(|f(x)-f(y)| = |x^2 - (x+\delta/2)^2| = |-x\delta - \delta^2/4| = |x\delta + \delta^2/4|\). We can make this value as large as we want by choosing a large \(x\). So, no single \(\delta\) works for all \(x\). Thus, \(x^2\) is not uniformly continuous on \(\mathbb{R}\).
- (C) Constant function \(f(x) = c\): This function is continuous. For any \(\epsilon > 0\), we can choose any \(\delta > 0\). If \(|x-y| < \delta\), then \(|f(x)-f(y)| = |c-c| = 0 < \epsilon\). So it is uniformly continuous.
- (D) Identity function \(f(x) = x\): This function is continuous. Its derivative is \(f'(x) = 1\), which is bounded. Therefore, it is uniformly continuous.
Step 4: Final Answer:
The function \(f(x)=x^2\) is continuous on \(\mathbb{R}\) but not uniformly continuous. This corresponds to option (B).