| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To determine if the function \(f(x) = \frac{x^3}{1 + x^2}\) is uniformly continuous on \(\mathbb{R}\), we need to check if for every \(\varepsilon > 0\), there exists a \(\delta > 0\) such that for all \(x, y \in \mathbb{R}\) with \(|x - y| < \delta\), we have \(|f(x) - f(y)| < \varepsilon\). The key to proving uniform continuity is often to show that the derivative of the function is bounded, as this implies that the function does not change too rapidly and thus can be uniformly continuous. Let's first find the derivative of \(f(x)\): \[ f'(x) = \frac{d}{dx}\left(\frac{x^3}{1 + x^2}\right) = \frac{(1 + x^2)(3x^2) - x^3(2x)}{(1 + x^2)^2} = \frac{3x^2 + 3x^4 - 2x^4}{(1 + x^2)^2} = \frac{3x^2 + x^4}{(1 + x^2)^2} \] To determine if \(f'(x)\) is bounded, we can analyze its behavior. Notice that the denominator \((1 + x^2)^2\) grows faster than the numerator \(3x^2 + x^4\) as \(x\) approaches \(\pm\infty\). This implies that \(\lim_{x \to \pm\infty} f'(x) = 0\), and since \(f'(x)\) is a continuous function, it is bounded on any finite interval. However, to show that \(f'(x)\) is bounded on \(\mathbb{R}\), we need to show that it does not grow without bound as \(x\) approaches \(\pm\infty\). Let's further analyze \(f'(x)\) by considering its behavior as \(x\) approaches \(\pm\infty\): \[ \lim_{x \to \pm\infty} f'(x) = \lim_{x \to \pm\infty} \frac{3x^2 + x^4}{(1 + x^2)^2} \] As \(x\) approaches \(\pm\infty\), the terms with the highest power of \(x\) dominate, so we can approximate the limit as: \[ \lim_{x \to \pm\infty} \frac{x^4}{x^4} = 1 \] Thus, not only does the limit exist, but it is also finite, indicating that \(f'(x)\) is bounded as \(x\) approaches \(\pm\infty\). Since \(f'(x)\) is continuous and bounded on \(\mathbb{R}\), \(f(x)\) is uniformly continuous on \(\mathbb{R}\). In conclusion, the function \(f(x) = \frac{x^3}{1 + x^2}\) is uniformly continuous on \(\mathbb{R}\). |
True |
| deepseekr1 | To determine if the function \( f(x) = \frac{x^3}{1 + x^2} \) is uniformly continuous on \(\mathbb{R}\), we analyze its derivative to check if it is bounded. A function with a bounded derivative is Lipschitz continuous, which implies uniform continuity. First, we compute the derivative \( f'(x) \) using the quotient rule: \[ f'(x) = \frac{(3x^2)(1 + x^2) - x^3(2x)}{(1 + x^2)^2} = \frac{3x^2 + 3x^4 - 2x^4}{(1 + x^2)^2} = \frac{3x^2 + x^4}{(1 + x^2)^2} \] Simplifying the expression, we rewrite the derivative as: \[ f'(x) = \frac{x^2(x^2 + 3)}{(1 + x^2)^2} \] To check if this derivative is bounded, we substitute \( t = x^2 \) (where \( t \geq 0 \)): \[ f'(x) = \frac{t(t + 3)}{(1 + t)^2} \] We analyze the function \( g(t) = \frac{t(t + 3)}{(1 + t)^2} \). By finding the critical points, we determine the maximum value of \( g(t) \). Solving for the critical points, we find that the maximum occurs at \( t = 3 \), giving: \[ g(3) = \frac{3 \cdot 6}{(1 + 3)^2} = \frac{18}{16} = \frac{9}{8} \] Thus, the maximum value of the derivative \( f'(x) \) is \( \frac{9}{8} \), which is bounded. Since the derivative is bounded, \( f \) is Lipschitz continuous, and therefore uniformly continuous on \(\mathbb{R}\). \[ \boxed{Yes} \] |
True |
| gpto1 | To determine whether the function \( f(x) = \dfrac{x^3}{1 + x^2} \) is uniformly continuous on \( \mathbb{R} \), we'll analyze the behavior of \( f \) and its derivative \( f' \) across its domain. **1. Understanding Uniform Continuity:** A function \( f: \mathbb{R} \rightarrow \mathbb{R} \) is uniformly continuous on \( \mathbb{R} \) if, for every \( \varepsilon > 0 \), there exists a \( \delta > 0 \) such that for all \( x, y \in \mathbb{R} \): \[ |x - y| < \delta \implies |f(x) - f(y)| < \varepsilon. \] The critical aspect of uniform continuity is that \( \delta \) is independent of \( x \) and \( y \); it depends only on \( \varepsilon \). **2. Behavior of \( f(x) \) at Infinity:** For large values of \( x \), we can approximate: \[ f(x) = \dfrac{x^3}{1 + x^2} = x \cdot \dfrac{x^2}{1 + x^2} \approx x \cdot 1 = x. \] Thus, as \( x \rightarrow \pm \infty \), \( f(x) \approx x \). **3. Exploring the Derivative \( f'(x) \):** Calculating the derivative will help us understand the rate at which \( f(x) \) changes. Using the quotient rule: \[ f'(x) = \dfrac{u'v - uv'}{v^2}, \] where \( u = x^3 \) and \( v = 1 + x^2 \). Compute \( u' \) and \( v' \): \[ u' = 3x^2, \quad v' = 2x. \] Compute numerator: \[ \begin{align*} u'v - uv' &= 3x^2(1 + x^2) - x^3(2x) \\ &= 3x^2(1 + x^2) - 2x^4 \\ &= 3x^2 + 3x^4 - 2x^4 \\ &= 3x^2 + x^4. \end{align*} \] Compute denominator: \[ v^2 = (1 + x^2)^2 = x^4 + 2x^2 + 1. \] Therefore: \[ f'(x) = \dfrac{3x^2 + x^4}{x^4 + 2x^2 + 1}. \] **4. Analyzing \( f'(x) \):** - **Limit as \( x \rightarrow 0 \):** At \( x = 0 \): \[ f'(0) = \dfrac{0 + 0}{0 + 0 + 1} = 0. \] - **Limit as \( x \rightarrow \infty \):** For large \( x \), the highest-degree terms dominate: \[ \begin{align*} \lim_{x \rightarrow \infty} f'(x) &= \lim_{x \rightarrow \infty} \dfrac{x^4 + 3x^2}{x^4 + 2x^2 + 1} \\ &= \lim_{x \rightarrow \infty} \dfrac{x^4}{x^4} = 1. \end{align*} \] - **Behavior on \( \mathbb{R} \):** For all \( x \in \mathbb{R} \), \( f'(x) \) is continuous and bounded between \( 0 \) and \( 1 \): \[ 0 \leq f'(x) \leq 1. \] **5. Implications for Uniform Continuity:** - **Lipschitz Continuity:** Since \( f' \) is bounded on \( \mathbb{R} \), \( f \) is Lipschitz continuous. Specifically, for all \( x, y \in \mathbb{R} \): \[ |f(x) - f(y)| = \left| \int_{y}^{x} f'(t) \, dt \right| \leq \int_{y}^{x} |f'(t)| \, dt \leq |x - y| \cdot \sup_{t \in \mathbb{R}} |f'(t)| \leq |x - y|. \] Thus, \( f \) is Lipschitz continuous with Lipschitz constant \( K = 1 \). - **Conclusion:** Lipschitz continuity implies uniform continuity. Therefore, \( f \) is uniformly continuous on \( \mathbb{R} \). **Answer: Yes; since its derivative is bounded, f is uniformly continuous on ℝ. |
True |
| GPT-4o | To determine whether the function \( f : \mathbb{R} \to \mathbb{R} \) defined by \( f(x) = \frac{x^3}{1 + x^2} \) is uniformly continuous on \( \mathbb{R} \), we need to analyze the behavior of \( f(x) \) as \( x \) varies over the real numbers and check if, for every \( \epsilon > 0 \), there exists a \( \delta > 0 \) such that for all \( x, y \in \mathbb{R} \) with \( |x - y| < \delta \), it holds that \( |f(x) - f(y)| < \epsilon \). ### Step-by-Step Solution: 1. **Function Analysis:** Consider the function \( f(x) = \frac{x^3}{1 + x^2} \). 2. **Behavior at Extreme Values:** - As \( x \to \infty \) or \( x \to -\infty \), the dominant term in the numerator \( x^3 \) and the term \( x^2 \) in the denominator suggest that the function behaves like \( x \). Specifically, \[ \lim_{x \to \infty} \frac{x^3}{1 + x^2} = \lim_{x \to \infty} \frac{x^3}{x^2 \left( \frac{1}{x^2} + 1 \right)} = \lim_{x \to \infty} \frac{x^3}{x^2} \cdot \frac{1}{1 + \frac{1}{x^2}} = \lim_{x \to \infty} x \cdot 1 = \infty. \] - Similarly, \[ \lim_{x \to -\infty} \frac{x^3}{1 + x^2} = -\infty. \] 3. **Checking Uniform Continuity:** - To check uniform continuity, consider the difference \( |f(x) - f(y)| \): \[ |f(x) - f(y)| = \left| \frac{x^3}{1 + x^2} - \frac{y^3}{1 + y^2} \right|. \] - For small values of \( x \) and \( y \) (say, within a bounded interval), the function \( f(x) \) may behave nicely. However, we need to check the behavior as \( x \) and \( y \) become large, where the function \( f(x) \) starts to approximate \( x \). - Consider \( x \) and \( y \) such that \( x \) is large and \( y \) is slightly smaller than \( x \). We approximate: \[ f(x) \approx x \quad \text{and} \quad f(y) \approx y, \] so that: \[ |f(x) - f(y)| \approx |x - y|. \] - If \( |x - y| \) is small, but \( x \) and \( y \) are very large, then \( |f(x) - f(y)| \) can still be large, which would violate the requirement for uniform continuity. 4. **Conclusion:** Since \( f(x) \approx x \) for very large \( x \), the difference \( |f(x) - f(y)| \approx |x - y| \) does not get arbitrarily small unless \( |x - y| \) is also very small, and this dependency on the magnitude of \( x \) and \( y \) suggests that \( f \) is not uniformly continuous. ### Final Answer: The function \( f(x) = \frac{x^3}{1 + x^2} \) is **not uniformly continuous** on \( \mathbb{R} \). |
True |
| llama3.1-70B-Instruct | To determine if f(x) = (x^3) / (1 + x^2) is uniformly continuous on R, we need to show that for every ε > 0, there exists a δ > 0 such that for all x, y in R, |x - y| < δ implies |f(x) - f(y)| < ε. First, let's rewrite f(x) as f(x) = x^3 / (1 + x^2) = x / (1 + 1/x^2). Now, let's consider the difference |f(x) - f(y)|: |f(x) - f(y)| = |x / (1 + 1/x^2) - y / (1 + 1/y^2)| Using the triangle inequality, we can bound this by: |f(x) - f(y)| ≤ |x - y| / (1 + 1/x^2) + |y| / (1 + 1/y^2) \* |1/x^2 - 1/y^2| Now, we can use the fact that 1 + 1/x^2 ≥ 1 for all x in R, so: |f(x) - f(y)| ≤ |x - y| + |y| / (1 + 1/y^2) \* |1/x^2 - 1/y^2| Next, we can use the fact that |1/x^2 - 1/y^2| = |(x^2 - y^2) / (x^2 y^2)| ≤ |x - y| / (x^2 y^2) ≤ |x - y| (since x^2 y^2 ≥ 1 for all x, y in R). So, we have: |f(x) - f(y)| ≤ |x - y| + |y| / (1 + 1/y^2) \* |x - y| Now, we can choose δ = ε / (1 + |y| / (1 + 1/y^2)). Then, if |x - y| < δ, we have: |f(x) - f(y)| ≤ |x - y| + |y| / (1 + 1/y^2) \* |x - y| < δ + |y| / (1 + 1/y^2) \* δ = ε Therefore, f(x) = (x^3) / (1 + x^2) is uniformly continuous on R. |
True |
| llama3.1-405B-Instruct | To determine if the function f(x) = (x^3) / (1 + x^2) is uniformly continuous on R, we need to show that for every ε > 0, there exists a δ > 0 such that for all x, y in R, |x - y| < δ implies |f(x) - f(y)| < ε. First, let's analyze the function f(x). We can rewrite it as: f(x) = (x^3) / (1 + x^2) = x / (1 + x^2) + x^3 / (1 + x^2) Now, let's consider the numerator and denominator separately: * The numerator x^3 is a continuous function on R, and it is bounded by |x|^3. * The denominator 1 + x^2 is also a continuous function on R, and it is bounded below by 1. Using these bounds, we can estimate the difference |f(x) - f(y)| as follows: |f(x) - f(y)| = |(x^3) / (1 + x^2) - (y^3) / (1 + y^2)| = |(x^3 - y^3) / (1 + x^2) + (y^3) / (1 + y^2) - (y^3) / (1 + x^2)| ≤ |(x^3 - y^3) / (1 + x^2)| + |(y^3) / (1 + y^2) - (y^3) / (1 + x^2)| Now, we can use the triangle inequality and the bounds on the numerator and denominator to further estimate the difference: |(x^3 - y^3) / (1 + x^2)| ≤ |x^3 - y^3| / (1 + x^2) ≤ |x - y| \* (|x|^2 + |x| \* |y| + |y|^2) / (1 + x^2) Similarly, |(y^3) / (1 + y^2) - (y^3) / (1 + x^2)| ≤ |y^3| \* |(1 + y^2) - (1 + x^2)| / (1 + x^2) ≤ |y|^3 \* |y^2 - x^2| / (1 + x^2) Combining these estimates, we get: |f(x) - f(y)| ≤ |x - y| \* (|x|^2 + |x| \* |y| + |y|^2) / (1 + x^2) + |y|^3 \* |y^2 - x^2| / (1 + x^2) Now, we can choose δ > 0 such that |x - y| < δ implies: |x|^2 + |x| \* |y| + |y|^2 < ε / (1 + x^2) |y|^3 \* |y^2 - x^2| < ε / (1 + x^2) This is possible because the expressions on the left-hand side are continuous functions of x and y, and they tend to 0 as x and y tend to infinity. Therefore, we have shown that for every ε > 0, there exists a δ > 0 such that for all x, y in R, |x - y| < δ implies |f(x) - f(y)| < ε. This means that the function f(x) = (x^3) / (1 + x^2) is uniformly continuous on R. |
True |
| DeepSeek-Coder-V2 | To determine if the function \( f(x) = \frac{x^3}{1 + x^2} \) is uniformly continuous on \( \mathbb{R} \), we need to show that for any \( \epsilon > 0 \), there exists a \( \delta > 0 \) such that for any \( x, y \in \mathbb{R} \) with \( |x - y| < \delta \), we have \( |f(x) - f(y)| < \epsilon \). First, let's analyze the function \( f(x) \). We can rewrite \( f(x) \) as: \[ f(x) = x \cdot \frac{x^2}{1 + x^2} = x \cdot \left(1 - \frac{1}{1 + x^2}\right) = x - \frac{x}{1 + x^2}. \] Now, consider the difference \( f(x) - f(y) \): \[ f(x) - f(y) = \left( x - \frac{x}{1 + x^2} \right) - \left( y - \frac{y}{1 + y^2} \right). \] Simplifying this, we get: \[ f(x) - f(y) = (x - y) - \left( \frac{x}{1 + x^2} - \frac{y}{1 + y^2} \right). \] To bound the second term, we need to estimate \( \left| \frac{x}{1 + x^2} - \frac{y}{1 + y^2} \right| \). Notice that: \[ \left| \frac{x}{1 + x^2} - \frac{y}{1 + y^2} \right| \leq \left| \frac{x}{1 + x^2} \right| + \left| \frac{y}{1 + y^2} \right|. \] Since \( \frac{x}{1 + x^2} \) and \( \frac{y}{1 + y^2} \) are both bounded by 1 (because \( \frac{|x|}{1 + x^2} \leq 1 \) and \( \frac{|y|}{1 + y^2} \leq 1 \)), we have: \[ \left| \frac{x}{1 + x^2} - \frac{y}{1 + y^2} \right| \leq 1 + 1 = 2. \] Thus, we can write: \[ |f(x) - f(y)| \leq |x - y| + 2 \left| \frac{x}{1 + x^2} - \frac{y}{1 + y^2} \right|. \] Given \( \epsilon > 0 \), we can choose \( \delta = \frac{\epsilon}{3} \). Then, for any \( x, y \in \mathbb{R} \) with \( |x - y| < \delta \), we have: \[ |f(x) - f(y)| \leq |x - y| + 2 \left| \frac{x}{1 + x^2} - \frac{y}{1 + y^2} \right| < \frac{\epsilon}{3} + 2 \cdot \frac{\epsilon}{3} = \epsilon. \] Therefore, \( f(x) \) is uniformly continuous on \( \mathbb{R} \). |
True |
| Qwen2-72B-Instruct | To determine if the function \(f(x) = \frac{x^3}{1 + x^2}\) is uniformly continuous on \(\mathbb{R}\), we need to check if for every \(\varepsilon > 0\), there exists a \(\delta > 0\) such that for all \(x, y \in \mathbb{R}\) with \(|x - y| < \delta\), we have \(|f(x) - f(y)| < \varepsilon\). The key to proving uniform continuity is often to show that the derivative of the function is bounded, as this implies that the function does not change too rapidly and thus can be uniformly continuous. Let's first find the derivative of \(f(x)\): \[ f'(x) = \frac{d}{dx}\left(\frac{x^3}{1 + x^2}\right) = \frac{(1 + x^2)(3x^2) - x^3(2x)}{(1 + x^2)^2} = \frac{3x^2 + 3x^4 - 2x^4}{(1 + x^2)^2} = \frac{3x^2 + x^4}{(1 + x^2)^2} \] To determine if \(f'(x)\) is bounded, we can analyze its behavior. Notice that the denominator \((1 + x^2)^2\) grows faster than the numerator \(3x^2 + x^4\) as \(x\) approaches \(\pm\infty\). This implies that \(\lim_{x \to \pm\infty} f'(x) = 0\), and since \(f'(x)\) is a continuous function, it is bounded on any finite interval. However, to show that \(f'(x)\) is bounded on \(\mathbb{R}\), we need to show that it does not grow without bound as \(x\) approaches \(\pm\infty\). Let's further analyze \(f'(x)\) by considering its behavior as \(x\) approaches \(\pm\infty\): \[ \lim_{x \to \pm\infty} f'(x) = \lim_{x \to \pm\infty} \frac{3x^2 + x^4}{(1 + x^2)^2} \] As \(x\) approaches \(\pm\infty\), the terms with the highest power of \(x\) dominate, so we can approximate the limit as: \[ \lim_{x \to \pm\infty} \frac{x^4}{x^4} = 1 \] Thus, not only does the limit exist, but it is also finite, indicating that \(f'(x)\) is bounded as \(x\) approaches \(\pm\infty\). Since \(f'(x)\) is continuous and bounded on \(\mathbb{R}\), \(f(x)\) is uniformly continuous on \(\mathbb{R}\). In conclusion, the function \(f(x) = \frac{x^3}{1 + x^2}\) is uniformly continuous on \(\mathbb{R}\). |
True |
| baichuan4 | To determine if the function \(f(x) = \frac{x^3}{1 + x^2}\) is uniformly continuous on \(\mathbb{R}\), we can use the following definition of uniform continuity: A function \(f\) is uniformly continuous on \(\mathbb{R}\) if for every \(\epsilon > 0\), there exists a \(\delta > 0\) such that for all \(x, y \in \mathbb{R}\), if \(|x - y| < \delta\), then \(|f(x) - f(y)| < \epsilon\). We can start by finding an expression for \(|f(x) - f(y)|\): \[|f(x) - f(y)| = \left|\frac{x^3}{1 + x^2} - \frac{y^3}{1 + y^2}\right|\] \[= \left|\frac{x^3(1 + y^2) - y^3(1 + x^2)}{(1 + x^2)(1 + y^2)}\right|\] \[= \left|\frac{x^3 + x^3y^2 - y^3 - x^2y^3}{(1 + x^2)(1 + y^2)}\right|\] \[= \left|\frac{(x^3 - y^3) + (x^3y^2 - x^2y^3)}{(1 + x^2)(1 + y^2)}\right|\] \[= \left|\frac{(x - y)(x^2 + xy + y^2) + xy(x - y)(x + y)}{(1 + x^2)(1 + y^2)}\right|\] \[= \left|\frac{(x - y)(x^2 + xy + y^2 + x^2y + xy^2)}{(1 + x^2)(1 + y^2)}\right|\] \[= |x - y|\left|\frac{x^2 + xy + y^2 + x^2y + xy^2}{(1 + x^2)(1 + y^2)}\right|\] Now, let's set \(\epsilon > 0\) and find a \(\delta > 0\) such that if \(|x - y| < \delta\), then \(|f(x) - f(y)| < \epsilon\). Looking at the expression we derived, we can see that the term \(\left|\frac{x^2 + xy + y^2 + x^2y + xy^2}{(1 + x^2)(1 + y^2)}\right|\) is bounded for all \(x, y \in \mathbb{R}\) because the denominator is always positive and the numerator is a polynomial in \(x\) and \(y\). Let's denote this bound by \(M\). Then, we have: \[|f(x) - f(y)| \leq M|x - y|\] So, if we choose \(\delta = \frac{\epsilon}{M}\), then whenever \(|x - y| < \delta\), it follows that \(|f(x) - f(y)| < \epsilon\). This shows that \(f(x) = \frac{x^3}{1 + x^2}\) is uniformly continuous on \(\mathbb{R}\) because for any \(\epsilon > 0\), we can find a \(\delta > 0\) that satisfies the definition of uniform continuity. |
True |
| doubao | To determine if f is uniformly continuous on R, we use the following theorem: If a function f is continuous on a closed and bounded interval [a, b], then it is uniformly continuous on [a, b]. First, we note that f is continuous on R since it is a quotient of continuous functions and the denominator is never zero. Now, consider any closed and bounded interval [a, b]. Then f is continuous on [a, b], and by the theorem, it is uniformly continuous on [a, b]. However, to show that f is uniformly continuous on the entire R, we need to show that for any ε > 0, there exists a δ > 0 such that for all x, y in R with |x - y| < δ, we have |f(x) - f(y)| < ε. Let ε > 0 be given. We have: |f(x) - f(y)| = |(x^3)/(1 + x^2) - (y^3)/(1 + y^2)| = |(x^3(1 + y^2) - y^3(1 + x^2))/((1 + x^2)(1 + y^2))| = |(x^3 + x^3y^2 - y^3 - y^3x^2)/((1 + x^2)(1 + y^2))| ≤ |x^3 - y^3| + |x^3y^2 - y^3x^2|/((1 + x^2)(1 + y^2)) ≤ |x - y|(|x^2 + xy + y^2|) + |xy(x^2 - y^2)|/((1 + x^2)(1 + y^2)) ≤ |x - y|(|x^2 + xy + y^2| + |xy||x + y|). Now, we can bound the terms |x^2 + xy + y^2| and |xy||x + y|. For example, |x^2 + xy + y^2| ≤ 3(max{|x|, |y|})^2 and |xy||x + y| ≤ |x|^2|y| + |y|^2|x| ≤ 2(max{|x|, |y|})^3. So, |f(x) - f(y)| ≤ |x - y|(3(max{|x|, |y|})^2 + 2(max{|x|, |y|})^3). To make this less than ε, we choose δ = ε/(3(max{|a|, |b|})^2 + 2(max{|a|, |b|})^3). Then for all x, y in R with |x - y| < δ, we have |f(x) - f(y)| < ε. Therefore, f is uniformly continuous on R. |
True |
| minimax | To determine if the function \( f(x) = \frac{x^3}{1 + x^2} \) is uniformly continuous on the real numbers \( \mathbb{R} \), we need to check if for every \( \epsilon > 0 \), there exists a \( \delta > 0 \) such that for all \( x, y \in \mathbb{R} \), if \( |x - y| < \delta \), then \( |f(x) - f(y)| < \epsilon \). A function is uniformly continuous on a set if it is continuous and its continuity does not depend on the location within the set. For polynomial functions and rational functions with non-zero denominators, we can often use the mean value theorem or other properties to determine uniform continuity. However, in this case, we can see that as \( x \) approaches infinity or negative infinity, the function \( f(x) \) behaves like \( x^3 \) because the denominator \( 1 + x^2 \) approaches infinity much slower than the numerator \( x^3 \). This means that the function is not bounded and will increase or decrease without bound as \( x \) goes to infinity or negative infinity. Since the function is not bounded, it cannot be uniformly continuous on the entire real line \( \mathbb{R} \). Uniform continuity would require that the function be bounded and that the rate of change of the function be controlled across the entire domain. However, as \( x \) becomes very large or very small, the function changes rapidly, which violates the condition for uniform continuity. Therefore, \( f(x) = \frac{x^3}{1 + x^2} \) is not uniformly continuous on \( \mathbb{R} \). |
True |
| kimi | To determine if the function \( f(x) = \frac{x^3}{1 + x^2} \) is uniformly continuous on \( \mathbb{R} \), we can use the definition of uniform continuity. A function is uniformly continuous on a set if for every \( \epsilon > 0 \), there exists a \( \delta > 0 \) such that for all \( x, y \in \mathbb{R} \), if \( |x - y| < \delta \), then \( |f(x) - f(y)| < \epsilon \). Let's consider the function \( f(x) \) and analyze its behavior. 1. **Behavior at Infinity**: As \( x \) approaches \( \pm \infty \), \( f(x) \) approaches 0 because the \( x^3 \) term is dominated by the \( x^2 \) term in the denominator. 2. **Derivative**: To check for uniform continuity, we can look at the derivative of \( f(x) \), which can give us an idea of how quickly \( f(x) \) changes. The derivative is given by: \[ f'(x) = \frac{d}{dx} \left( \frac{x^3}{1 + x^2} \right) = \frac{3x^2(1 + x^2) - x^3 \cdot 2x}{(1 + x^2)^2} = \frac{3x^2 + 3x^4 - 2x^4}{(1 + x^2)^2} = \frac{3x^2(1 - x^2 + x^2)}{(1 + x^2)^2} = \frac{3x^2}{(1 + x^2)^2} \] As \( x \) approaches \( \pm \infty \), \( f'(x) \) approaches 0. 3. **Lipschitz Condition**: A function is uniformly continuous if it satisfies a Lipschitz condition, which means there exists a constant \( L \) such that \( |f(x) - f(y)| \leq L|x - y| \) for all \( x, y \). The derivative \( f'(x) \) gives us a bound on the rate of change of \( f(x) \), but since \( f'(x) \) approaches 0 as \( x \) approaches infinity, we cannot find a constant \( L \) that works for all \( x \). 4. **Counterexample**: To show that \( f(x) \) is not uniformly continuous, we can provide a counterexample. Consider \( x_n = \sqrt{n} \) and \( y_n = \sqrt{n} + \frac{1}{n} \). Then \( |x_n - y_n| = \frac{1}{n} \), but: \[ |f(x_n) - f(y_n)| = \left| \frac{\sqrt{n}^3}{1 + \sqrt{n}^2} - \frac{(\sqrt{n} + \frac{1}{n})^3}{1 + (\sqrt{n} + \frac{1}{n})^2} \right| \] As \( n \) approaches infinity, the second term in the numerator of the second fraction becomes negligible compared to the first term, so the difference does not approach 0 as fast as \( \frac{1}{n} \), which means \( f(x) \) is not uniformly continuous. Therefore, the function \( f(x) = \frac{x^3}{1 + x^2} \) is not uniformly continuous on \( \mathbb{R} \). |
True |
| wenxinyiyan | To determine if the function $f(x) = \frac{x^3}{1 + x^2}$ is uniformly continuous on $\mathbb{R}$, we can use the following steps: Step 1: Show that $f(x)$ is continuous on $\mathbb{R}$. The function $f(x)$ is a rational function, and its denominator $1 + x^2$ is never zero for any real number $x$. Therefore, $f(x)$ is continuous on $\mathbb{R}$. Step 2: Analyze the behavior of $f(x)$ as $x$ approaches infinity. As $x$ approaches infinity, the term $x^2$ dominates the denominator, and $f(x)$ behaves like $x$. More precisely, we can show that $f(x)$ is not uniformly continuous by finding a counterexample to the uniform continuity condition. Step 3: Find a counterexample to uniform continuity. Consider the sequences $x_n = n$ and $y_n = n + \frac{1}{n}$ for $n \in \mathbb{N}$. Then, $$ \lim_{{n \to \infty}} (x_n - y_n) = \lim_{{n \to \infty}} \left(n - (n + \frac{1}{n})\right) = \lim_{{n \to \infty}} -\frac{1}{n} = 0. $$ However, $$ \lim_{{n \to \infty}} (f(x_n) - f(y_n)) = \lim_{{n \to \infty}} \left(\frac{n^3}{1 + n^2} - \frac{(n + \frac{1}{n})^3}{1 + (n + \frac{1}{n})^2}\right) = \lim_{{n \to \infty}} \left(\frac{n^3}{1 + n^2} - \frac{n^3 + 3n + \frac{1}{n^3}}{1 + 2n^2 + \frac{1}{n^2}}\right) = 1 - 1 = 0. $$ Since the difference in function values does not approach zero as the difference in the inputs approaches zero, $f(x)$ is not uniformly continuous on $\mathbb{R}$. Note: The counterexample provided in Step 3 is incorrect. Let's correct it. Consider the sequences $x_n = n$ and $y_n = n + 1$ for $n \in \mathbb{N}$. Then, $$ \lim_{{n \to \infty}} (x_n - y_n) = \lim_{{n \to \infty}} (n - (n + 1)) = \lim_{{n \to \infty}} -1 = -1. $$ However, $$ \lim_{{n \to \infty}} (f(x_n) - f(y_n)) = \lim_{{n \to \infty}} \left(\frac{n^3}{1 + n^2} - \frac{(n + 1)^3}{1 + (n + 1)^2}\right) = \lim_{{n \to \infty}} \left(\frac{n^3}{1 + n^2} - \frac{n^3 + 3n^2 + 3n + 1}{1 + 2n^2 + 2n + 1}\right) = 1 - 1 = 0. $$ Since the difference in function values does not approach zero as the difference in the inputs approaches zero, $f(x)$ is not uniformly continuous on $\mathbb{R}$. Note: The corrected counterexample is also incorrect. Let's provide a correct counterexample. Consider the sequences $x_n = n$ and $y_n = n + \frac{1}{n}$ for $n \in \mathbb{N}$. Then, $$ \lim_{{n \to \infty}} (x_n - y_n) = \lim_{{n \to \infty}} \left(n - (n + \frac{1}{n})\right) = \lim_{{n \to \infty}} -\frac{1}{n} = 0. $$ However, $$ \lim_{{n \to \infty}} (f(x_n) - f(y_n)) = \lim_{{n \to \infty}} \left(\frac{n^3}{1 + n^2} - \frac{(n + \frac{1}{n})^3}{1 + (n + \frac{1}{n |
True |
| zhipu | To determine if the function \( f(x) = \frac{x^3}{1 + x^2} \) is uniformly continuous on \( \mathbb{R} \), we need to show that for every \( \epsilon > 0 \), there exists a \( \delta > 0 \) such that for all \( x, y \in \mathbb{R} \), if \( |x - y| < \delta \), then \( |f(x) - f(y)| < \epsilon \). ### Step-by-Step Solution: 1. **Expression for \( |f(x) - f(y)| \)**: \[ |f(x) - f(y)| = \left| \frac{x^3}{1 + x^2} - \frac{y^3}{1 + y^2} \right| \] 2. **Common Denominator**: \[ |f(x) - f(y)| = \left| \frac{x^3(1 + y^2) - y^3(1 + x^2)}{(1 + x^2)(1 + y^2)} \right| \] \[ = \left| \frac{x^3 + x^3 y^2 - y^3 - y^3 x^2}{(1 + x^2)(1 + y^2)} \right| \] \[ = \left| \frac{x^3 - y^3 + x^3 y^2 - y^3 x^2}{(1 + x^2)(1 + y^2)} \right| \] \[ = \left| \frac{(x - y)(x^2 + xy + y^2) + xy(x - y)(x + y)}{(1 + x^2)(1 + y^2)} \right| \] \[ = \left| \frac{(x - y)(x^2 + xy + y^2 + xy(x + y))}{(1 + x^2)(1 + y^2)} \right| \] \[ = |x - y| \left| \frac{x^2 + xy + y^2 + xy(x + y)}{(1 + x^2)(1 + y^2)} \right| \] 3. **Analyzing the Expression**: Notice that the term \( \left| \frac{x^2 + xy + y^2 + xy(x + y)}{(1 + x^2)(1 + y^2)} \right| \) is not bounded as \( x \) and \( y \) go to infinity. Specifically, as \( x \) and \( y \) become very large, the numerator \( x^2 + xy + y^2 + xy(x + y) \) grows faster than the denominator \( (1 + x^2)(1 + y^2) \). 4. **Conclusion**: Because the expression \( \left| \frac{x^2 + xy + y^2 + xy(x + y)}{(1 + x^2)(1 + y^2)} \right| \) is unbounded, there does not exist a constant \( M \) such that \( |f(x) - f(y)| \leq M |x - y| \) for all \( x, y \in \mathbb{R} \). This implies that \( f \) is not uniformly continuous on \( \mathbb{R} \). ### Alternative Approach: Another way to see this is to consider the behavior of \( f(x) \) as \( x \to \infty \). The derivative \( f'(x) \) can be used to analyze the uniform continuity: \[ f'(x) = \frac{(3x^2)(1 + x^2) - x^3(2x)}{(1 + x^2)^2} = \frac{3x^2 + 3x^4 - 2x^4}{(1 + x^2)^2} = \frac{x^2(3 + x^2 - 2x^2)}{(1 + x^2)^2} = \frac{x^2(1 + x^2)}{(1 + x^2)^2} = \frac{x^2}{1 + x^2} \] As \( x \to \infty \), \( f'(x) \to 1 \). Since \( f'(x) \) does not tend to zero, \( f(x) \) is not uniformly continuous. ### Final Conclusion: The function \( f(x) = \frac{x^3}{1 + x^2} \) is **not uniformly continuous** on \( \mathbb{R} \). |
True |
| gpto1 | To determine whether the function \( f(x) = \dfrac{x^3}{1 + x^2} \) is uniformly continuous on \( \mathbb{R} \), we'll analyze the behavior of \( f \) and its derivative \( f' \) across its domain. **1. Understanding Uniform Continuity:** A function \( f: \mathbb{R} \rightarrow \mathbb{R} \) is uniformly continuous on \( \mathbb{R} \) if, for every \( \varepsilon > 0 \), there exists a \( \delta > 0 \) such that for all \( x, y \in \mathbb{R} \): \[ |x - y| < \delta \implies |f(x) - f(y)| < \varepsilon. \] The critical aspect of uniform continuity is that \( \delta \) is independent of \( x \) and \( y \); it depends only on \( \varepsilon \). **2. Behavior of \( f(x) \) at Infinity:** For large values of \( x \), we can approximate: \[ f(x) = \dfrac{x^3}{1 + x^2} = x \cdot \dfrac{x^2}{1 + x^2} \approx x \cdot 1 = x. \] Thus, as \( x \rightarrow \pm \infty \), \( f(x) \approx x \). **3. Exploring the Derivative \( f'(x) \):** Calculating the derivative will help us understand the rate at which \( f(x) \) changes. Using the quotient rule: \[ f'(x) = \dfrac{u'v - uv'}{v^2}, \] where \( u = x^3 \) and \( v = 1 + x^2 \). Compute \( u' \) and \( v' \): \[ u' = 3x^2, \quad v' = 2x. \] Compute numerator: \[ \begin{align*} u'v - uv' &= 3x^2(1 + x^2) - x^3(2x) \\ &= 3x^2(1 + x^2) - 2x^4 \\ &= 3x^2 + 3x^4 - 2x^4 \\ &= 3x^2 + x^4. \end{align*} \] Compute denominator: \[ v^2 = (1 + x^2)^2 = x^4 + 2x^2 + 1. \] Therefore: \[ f'(x) = \dfrac{3x^2 + x^4}{x^4 + 2x^2 + 1}. \] **4. Analyzing \( f'(x) \):** - **Limit as \( x \rightarrow 0 \):** At \( x = 0 \): \[ f'(0) = \dfrac{0 + 0}{0 + 0 + 1} = 0. \] - **Limit as \( x \rightarrow \infty \):** For large \( x \), the highest-degree terms dominate: \[ \begin{align*} \lim_{x \rightarrow \infty} f'(x) &= \lim_{x \rightarrow \infty} \dfrac{x^4 + 3x^2}{x^4 + 2x^2 + 1} \\ &= \lim_{x \rightarrow \infty} \dfrac{x^4}{x^4} = 1. \end{align*} \] - **Behavior on \( \mathbb{R} \):** For all \( x \in \mathbb{R} \), \( f'(x) \) is continuous and bounded between \( 0 \) and \( 1 \): \[ 0 \leq f'(x) \leq 1. \] **5. Implications for Uniform Continuity:** - **Lipschitz Continuity:** Since \( f' \) is bounded on \( \mathbb{R} \), \( f \) is Lipschitz continuous. Specifically, for all \( x, y \in \mathbb{R} \): \[ |f(x) - f(y)| = \left| \int_{y}^{x} f'(t) \, dt \right| \leq \int_{y}^{x} |f'(t)| \, dt \leq |x - y| \cdot \sup_{t \in \mathbb{R}} |f'(t)| \leq |x - y|. \] Thus, \( f \) is Lipschitz continuous with Lipschitz constant \( K = 1 \). - **Conclusion:** Lipschitz continuity implies uniform continuity. Therefore, \( f \) is uniformly continuous on \( \mathbb{R} \). **Answer: Yes; since its derivative is bounded, f is uniformly continuous on ℝ. |
True |
| deepseekv3 | To determine whether the function \( f: \mathbb{R} \to \mathbb{R} \) defined by \( f(x) = \frac{x^3}{1 + x^2} \) is uniformly continuous on \( \mathbb{R} \), we can proceed with the following analysis. --- ### Step 1: Compute the derivative of \( f(x) \) To analyze the behavior of \( f(x) \), we first compute its derivative using the quotient rule: \[ f'(x) = \frac{d}{dx} \left( \frac{x^3}{1 + x^2} \right) = \frac{(3x^2)(1 + x^2) - x^3(2x)}{(1 + x^2)^2}. \] Simplify the numerator: \[ 3x^2(1 + x^2) - 2x^4 = 3x^2 + 3x^4 - 2x^4 = 3x^2 + x^4. \] Thus, the derivative is: \[ f'(x) = \frac{3x^2 + x^4}{(1 + x^2)^2}. \] --- ### Step 2: Analyze the behavior of \( f'(x) \) To determine whether \( f(x) \) is uniformly continuous, we need to examine whether \( f'(x) \) is bounded on \( \mathbb{R} \). If \( f'(x) \) is bounded, then \( f(x) \) is Lipschitz continuous, which implies uniform continuity. Rewrite \( f'(x) \) as: \[ f'(x) = \frac{x^4 + 3x^2}{(1 + x^2)^2}. \] Divide the numerator and denominator by \( x^4 \) (assuming \( x \neq 0 \)): \[ f'(x) = \frac{1 + \frac{3}{x^2}}{\left(\frac{1}{x^2} + 1\right)^2}. \] As \( |x| \to \infty \), \( \frac{1}{x^2} \to 0 \), so: \[ \lim_{|x| \to \infty} f'(x) = \frac{1 + 0}{(0 + 1)^2} = 1. \] For \( x = 0 \), we have \( f'(0) = 0 \). Thus, \( f'(x) \) is bounded on \( \mathbb{R} \), with: \[ |f'(x)| \leq 1 \quad \text{for all } x \in \mathbb{R}. \] --- ### Step 3: Conclude uniform continuity Since \( f'(x) \) is bounded on \( \mathbb{R} \), \( f(x) \) is Lipschitz continuous. Lipschitz continuity implies uniform continuity. Therefore, \( f(x) \) is uniformly continuous on \( \mathbb{R} \). --- ### Final Answer: Yes, \( f(x) = \frac{x^3}{1 + x^2} \) is uniformly continuous on \( \mathbb{R} \). |
True |