| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To estimate \((3.99)^3 \cdot (1.01)^4 \cdot (1.98)^{-1}\) using linear approximation, we can apply the concept of the differential to approximate each factor. The differential approximation for a function \(f(x)\) near \(x=a\) is given by: \[ f(x) \approx f(a) + f'(a)(x-a) \] where \(f'(a)\) is the derivative of \(f\) evaluated at \(x=a\). Let's apply this to each factor: 1. For \((3.99)^3\), we have \(f(x) = x^3\). Let \(a=4\), so \(x=3.99\), then \(f(4) = 4^3 = 64\) and \(f'(x) = 3x^2\), so \(f'(4) = 3 \cdot 4^2 = 48\). The approximation is: \[ (3.99)^3 \approx f(4) + f'(4)(3.99-4) = 64 + 48 \cdot (-0.01) = 64 - 0.48 = 63.52 \] 2. For \((1.01)^4\), we have \(f(x) = x^4\). Let \(a=1\), so \(x=1.01\), then \(f(1) = 1^4 = 1\) and \(f'(x) = 4x^3\), so \(f'(1) = 4 \cdot 1^3 = 4\). The approximation is: \[ (1.01)^4 \approx f(1) + f'(1)(1.01-1) = 1 + 4 \cdot (0.01) = 1 + 0.04 = 1.04 \] 3. For \((1.98)^{-1}\), we have \(f(x) = x^{-1}\). Let \(a=2\), so \(x=1.98\), then \(f(2) = 2^{-1} = 0.5\) and \(f'(x) = -x^{-2}\), so \(f'(2) = -2^{-2} = -0.25\). The approximation is: \[ (1.98)^{-1} \approx f(2) + f'(2)(1.98-2) = 0.5 - 0.25 \cdot (-0.02) = 0.5 + 0.005 = 0.505 \] Multiplying the approximations, we get: \[ (3.99)^3 \cdot (1.01)^4 \cdot (1.98)^{-1} \approx 63.52 \cdot 1.04 \cdot 0.505 \approx 33.21 \] Therefore, the estimate for \((3.99)^3 \cdot (1.01)^4 \cdot (1.98)^{-1}\) using linear approximation is approximately 33.21. |
33.36 |
| deepseekr1 | To estimate \((3.99)^3 (1.01)^4 (1.98)^{-1}\) using linear approximation, we approximate the function \(f(x, y, z) = \frac{x^3 y^4}{z}\) around the point \((4, 1, 2)\). 1. **Base Value Calculation**: \[ f(4, 1, 2) = \frac{4^3 \cdot 1^4}{2} = \frac{64}{2} = 32 \] 2. **Partial Derivatives**: - Partial derivative with respect to \(x\): \[ f_x = \frac{3x^2 y^4}{z} \quad \Rightarrow \quad f_x(4, 1, 2) = \frac{3 \cdot 16 \cdot 1}{2} = 24 \] - Partial derivative with respect to \(y\): \[ f_y = \frac{4x^3 y^3}{z} \quad \Rightarrow \quad f_y(4, 1, 2) = \frac{4 \cdot 64 \cdot 1}{2} = 128 \] - Partial derivative with respect to \(z\): \[ f_z = -\frac{x^3 y^4}{z^2} \quad \Rightarrow \quad f_z(4, 1, 2) = -\frac{64}{4} = -16 \] 3. **Linear Approximation**: - Changes in variables: \(\Delta x = -0.01\), \(\Delta y = 0.01\), \(\Delta z = -0.02\) - Using the linear approximation formula: \[ f(4 - 0.01, 1 + 0.01, 2 - 0.02) \approx f(4, 1, 2) + f_x \Delta x + f_y \Delta y + f_z \Delta z \] \[ \approx 32 + 24(-0.01) + 128(0.01) + (-16)(-0.02) \] \[ = 32 - 0.24 + 1.28 + 0.32 \] \[ = 32 + 1.36 = 33.36 \] Thus, the estimated value using linear approximation is \(\boxed{33.36}\). |
33.36 |
| gpto1 | To estimate the value of \((3.99)^3 (1.01)^4 (1.98)^{-1}\) using linear approximation, we'll employ logarithms and properties of exponentials for small deviations. **Step 1: Recognize Small Deviations** Notice that: - \(3.99\) is close to \(4\) - \(1.01\) is close to \(1\) - \(1.98\) is close to \(2\) Let: - \(h_1 = 3.99 - 4 = -0.01\) - \(h_2 = 1.01 - 1 = 0.01\) - \(h_3 = 1.98 - 2 = -0.02\) **Step 2: Take Natural Logarithms** Consider the natural logarithm of the expression: \[ \ln\left[(3.99)^3 (1.01)^4 (1.98)^{-1}\right] = 3\ln(3.99) + 4\ln(1.01) - \ln(1.98) \] **Step 3: Linearize Each Logarithm** For small \(h\), the linear approximation is \(\ln(a + h) \approx \ln(a) + \dfrac{h}{a}\). Compute each term: 1. **\(\ln(3.99)\):** \[ \ln(3.99) = \ln(4 - 0.01) \approx \ln(4) - \dfrac{0.01}{4} = \ln(4) - 0.0025 \] 2. **\(\ln(1.01)\):** \[ \ln(1.01) = \ln(1 + 0.01) \approx \ln(1) + 0.01 = 0 + 0.01 \] 3. **\(\ln(1.98)\):** \[ \ln(1.98) = \ln(2 - 0.02) \approx \ln(2) - \dfrac{0.02}{2} = \ln(2) - 0.01 \] **Step 4: Combine the Terms** Substitute back into the original expression: \[ \begin{align*} 3\ln(3.99) + 4\ln(1.01) - \ln(1.98) &\approx 3\left(\ln(4) - 0.0025\right) + 4(0.01) - \left(\ln(2) - 0.01\right) \\ &= 3\ln(4) - 0.0075 + 0.04 - \ln(2) + 0.01. \end{align*} \] Simplify using \(\ln(4) = 2\ln(2)\): \[ \begin{align*} 3\ln(4) - \ln(2) &= 3(2\ln(2)) - \ln(2) = (6\ln(2)) - \ln(2) = 5\ln(2). \end{align*} \] Combine constants: \[ -0.0075 + 0.04 + 0.01 = 0.0425. \] So the total logarithm is: \[ 5\ln(2) + 0.0425. \] **Step 5: Exponentiate to Find the Approximate Value** \[ (3.99)^3 (1.01)^4 (1.98)^{-1} \approx e^{5\ln(2) + 0.0425} = 2^5 e^{0.0425} = 32 \times e^{0.0425}. \] **Step 6: Approximate \(e^{0.0425}\)** For small \(x\), \(e^{x} \approx 1 + x + \dfrac{x^2}{2}\): \[ e^{0.0425} \approx 1 + 0.0425 + \dfrac{(0.0425)^2}{2} = 1 + 0.0425 + 0.000903125 = 1.043403125. \] **Step 7: Calculate the Final Approximation** Multiply: \[ 32 \times 1.043403125 \approx 33.3889. \] **Conclusion:** The estimated value of the expression is approximately \(33.39\). **Answer:** Approximately **33.39** |
33.36 |
| GPT-4o | To solve the problem \((3.99)^3 (1.01)^4 (1.98)^{-1}\) using linear approximation, we can use the fact that for a function \( f(x) \) and a point \( a \), the linear approximation (or tangent line approximation) of the function at \( a \) is given by: \[ f(x) \approx f(a) + f'(a)(x - a) \] Let's break it down into three separate functions and then combine them: 1. \( f_1(x) = x^3 \) with \( x = 3.99 \) and \( a = 4 \) 2. \( f_2(x) = x^4 \) with \( x = 1.01 \) and \( a = 1 \) 3. \( f_3(x) = x^{-1} \) with \( x = 1.98 \) and \( a = 2 \) ### Step 1: Linear Approximation for \( (3.99)^3 \) For \( f_1(x) = x^3 \): \[ f_1(a) = 4^3 = 64 \] \[ f_1'(x) = 3x^2 \] \[ f_1'(4) = 3 \cdot 4^2 = 48 \] So the linear approximation at \( x = 3.99 \) is: \[ f_1(3.99) \approx 64 + 48 (3.99 - 4) \] \[ f_1(3.99) \approx 64 + 48 \cdot (-0.01) \] \[ f_1(3.99) \approx 64 - 0.48 \] \[ f_1(3.99) \approx 63.52 \] ### Step 2: Linear Approximation for \( (1.01)^4 \) For \( f_2(x) = x^4 \): \[ f_2(a) = 1^4 = 1 \] \[ f_2'(x) = 4x^3 \] \[ f_2'(1) = 4 \cdot 1^3 = 4 \] So the linear approximation at \( x = 1.01 \) is: \[ f_2(1.01) \approx 1 + 4 (1.01 - 1) \] \[ f_2(1.01) \approx 1 + 4 \cdot 0.01 \] \[ f_2(1.01) \approx 1 + 0.04 \] \[ f_2(1.01) \approx 1.04 \] ### Step 3: Linear Approximation for \( (1.98)^{-1} \) For \( f_3(x) = x^{-1} \): \[ f_3(a) = 2^{-1} = \frac{1}{2} = 0.5 \] \[ f_3'(x) = -x^{-2} \] \[ f_3'(2) = -2^{-2} = -\frac{1}{4} = -0.25 \] So the linear approximation at \( x = 1.98 \) is: \[ f_3(1.98) \approx 0.5 - 0.25 (1.98 - 2) \] \[ f_3(1.98) \approx 0.5 - 0.25 \cdot (-0.02) \] \[ f_3(1.98) \approx 0.5 + 0.005 \] \[ f_3(1.98) \approx 0.505 \] ### Combining the Approximations Now we combine the approximations: \[ (3.99)^3 (1.01)^4 (1.98)^{-1} \approx 63.52 \cdot 1.04 \cdot 0.505 \] First, multiply \( 63.52 \) and \( 1.04 \): \[ 63.52 \cdot 1.04 \approx 66.0608 \] Next, multiply by \( 0.505 \): \[ 66.0608 \cdot 0.505 \approx 33.367304 \] Therefore, the linear approximation for \( (3.99)^3 (1.01)^4 (1.98)^{-1} \) is approximately: \[ \boxed{33.37} \] |
33.36 |
| llama3.1-70B-Instruct | To estimate the value of (3.99)^3 (1.01)^4 (1.98)^{-1} using linear approximation, we can start by finding the linear approximation of each term separately. 1. For (3.99)^3: Let f(x) = x^3. Then f'(x) = 3x^2. We want to find the linear approximation of f(x) at x = 4, since 3.99 is close to 4. f(4) = 4^3 = 64 f'(4) = 3(4)^2 = 48 The linear approximation of f(x) at x = 4 is: f(x) ≈ f(4) + f'(4)(x - 4) = 64 + 48(x - 4) Now, we can plug in x = 3.99: f(3.99) ≈ 64 + 48(3.99 - 4) = 64 + 48(-0.01) = 64 - 0.48 = 63.52 So, (3.99)^3 ≈ 63.52. 2. For (1.01)^4: Let f(x) = x^4. Then f'(x) = 4x^3. We want to find the linear approximation of f(x) at x = 1, since 1.01 is close to 1. f(1) = 1^4 = 1 f'(1) = 4(1)^3 = 4 The linear approximation of f(x) at x = 1 is: f(x) ≈ f(1) + f'(1)(x - 1) = 1 + 4(x - 1) Now, we can plug in x = 1.01: f(1.01) ≈ 1 + 4(1.01 - 1) = 1 + 4(0.01) = 1 + 0.04 = 1.04 So, (1.01)^4 ≈ 1.04. 3. For (1.98)^{-1}: Let f(x) = x^{-1}. Then f'(x) = -x^{-2}. We want to find the linear approximation of f(x) at x = 2, since 1.98 is close to 2. f(2) = 2^{-1} = 1/2 f'(2) = -2^{-2} = -1/4 The linear approximation of f(x) at x = 2 is: f(x) ≈ f(2) + f'(2)(x - 2) = 1/2 - 1/4(x - 2) Now, we can plug in x = 1.98: f(1.98) ≈ 1/2 - 1/4(1.98 - 2) = 1/2 - 1/4(-0.02) = 1/2 + 0.005 = 0.505 So, (1.98)^{-1} ≈ 0.505. Now, we can multiply the estimated values: (3.99)^3 (1.01)^4 (1.98)^{-1} ≈ 63.52 × 1.04 × 0.505 ≈ 33.51 Therefore, the estimated value of (3.99)^3 (1.01)^4 (1.98)^{-1} using linear approximation is approximately 33.51. |
33.36 |
| llama3.1-405B-Instruct | To estimate (3.99)^3 (1.01)^4 (1.98)^{-1} using linear approximation, we can start by finding the linear approximation of each factor. 1. (3.99)^3: Let f(x) = x^3. Then, f'(x) = 3x^2. The linear approximation of f(x) at x = 4 is: f(4 + h) ≈ f(4) + f'(4)h where h = -0.01 (since 3.99 = 4 - 0.01). f(4) = 4^3 = 64 f'(4) = 3(4)^2 = 48 So, f(3.99) ≈ 64 + 48(-0.01) = 64 - 0.48 = 63.52 2. (1.01)^4: Let g(x) = x^4. Then, g'(x) = 4x^3. The linear approximation of g(x) at x = 1 is: g(1 + h) ≈ g(1) + g'(1)h where h = 0.01 (since 1.01 = 1 + 0.01). g(1) = 1^4 = 1 g'(1) = 4(1)^3 = 4 So, g(1.01) ≈ 1 + 4(0.01) = 1 + 0.04 = 1.04 3. (1.98)^{-1}: Let h(x) = x^{-1}. Then, h'(x) = -x^{-2}. The linear approximation of h(x) at x = 2 is: h(2 + h) ≈ h(2) + h'(2)h where h = -0.02 (since 1.98 = 2 - 0.02). h(2) = 2^{-1} = 1/2 h'(2) = -2^{-2} = -1/4 So, h(1.98) ≈ 1/2 + (-1/4)(-0.02) = 1/2 + 0.005 = 0.505 Now, we can multiply the linear approximations of each factor to estimate the original expression: (3.99)^3 (1.01)^4 (1.98)^{-1} ≈ (63.52)(1.04)(0.505) ≈ 33.33 Therefore, the estimated value of (3.99)^3 (1.01)^4 (1.98)^{-1} using linear approximation is 33.33. |
33.36 |
| DeepSeek-Coder-V2 | To solve the problem of estimating \((3.99)^3 (1.01)^4 (1.98)^{-1}\) using linear approximation, we will follow these steps: 1. **Identify the functions and their approximations:** - For \(3.99\), we use \(f(x) = x^3\) and approximate around \(x = 4\). - For \(1.01\), we use \(g(x) = x^4\) and approximate around \(x = 1\). - For \(1.98\), we use \(h(x) = x^{-1}\) and approximate around \(x = 2\). 2. **Compute the linear approximations:** - For \(f(x) = x^3\) around \(x = 4\): \[ f(4) = 4^3 = 64 \] \[ f'(x) = 3x^2 \quad \Rightarrow \quad f'(4) = 3 \cdot 4^2 = 48 \] \[ f(3.99) \approx f(4) + f'(4) \cdot (3.99 - 4) = 64 + 48 \cdot (-0.01) = 64 - 0.48 = 63.52 \] - For \(g(x) = x^4\) around \(x = 1\): \[ g(1) = 1^4 = 1 \] \[ g'(x) = 4x^3 \quad \Rightarrow \quad g'(1) = 4 \cdot 1^3 = 4 \] \[ g(1.01) \approx g(1) + g'(1) \cdot (1.01 - 1) = 1 + 4 \cdot 0.01 = 1 + 0.04 = 1.04 \] - For \(h(x) = x^{-1}\) around \(x = 2\): \[ h(2) = 2^{-1} = 0.5 \] \[ h'(x) = -x^{-2} \quad \Rightarrow \quad h'(2) = -2^{-2} = -0.25 \] \[ h(1.98) \approx h(2) + h'(2) \cdot (1.98 - 2) = 0.5 + (-0.25) \cdot (-0.02) = 0.5 + 0.005 = 0.505 \] 3. **Combine the approximations:** \[ (3.99)^3 (1.01)^4 (1.98)^{-1} \approx 63.52 \cdot 1.04 \cdot 0.505 \] 4. **Calculate the final estimate:** \[ 63.52 \cdot 1.04 = 66.0608 \] \[ 66.0608 \cdot 0.505 = 33.360704 \] Thus, the estimated value of \((3.99)^3 (1.01)^4 (1.98)^{-1}\) using linear approximation is approximately \(\boxed{33.360704}\). |
33.36 |
| Qwen2-72B-Instruct | To estimate (3.99)^3 * (1.01)^4 * (1.98)^{-1} using linear approximation, we can apply the concept of the differential to approximate each factor. The differential approximation for a function f(x) near x=a is given by: f(x) ≈ f(a) + f'(a)(x-a) where f'(a) is the derivative of f evaluated at x=a. Let's apply this to each factor: 1. For (3.99)^3, we have f(x) = x^3. Let a=4, so x=3.99, then f(4) = 4^3 = 64 and f'(x) = 3x^2, so f'(4) = 3*4^2 = 48. The approximation is: (3.99)^3 ≈ f(4) + f'(4)(3.99-4) = 64 + 48*(-0.01) = 64 - 0.48 = 63.52 2. For (1.01)^4, we have f(x) = x^4. Let a=1, so x=1.01, then f(1) = 1^4 = 1 and f'(x) = 4x^3, so f'(1) = 4*1^3 = 4. The approximation is: (1.01)^4 ≈ f(1) + f'(1)(1.01-1) = 1 + 4*(0.01) = 1 + 0.04 = 1.04 3. For (1.98)^{-1}, we have f(x) = x^{-1}. Let a=2, so x=1.98, then f(2) = 2^{-1} = 0.5 and f'(x) = -x^{-2}, so f'(2) = -2^{-2} = -0.25. The approximation is: (1.98)^{-1} ≈ f(2) + f'(2)(1.98-2) = 0.5 - 0.25*(-0.02) = 0.5 + 0.005 = 0.505 Multiplying the approximations, we get: (3.99)^3 * (1.01)^4 * (1.98)^{-1} ≈ 63.52 * 1.04 * 0.505 ≈ 33.21 Therefore, the estimate for (3.99)^3 * (1.01)^4 * (1.98)^{-1} using linear approximation is approximately 33.21. |
33.36 |
| baichuan4 | To estimate $(3.99)^3 (1.01)^4 (1.98)^{-1}$ using linear approximation, we can start by taking the natural logarithm of the expression and then use the fact that for small changes in $x$, $\ln(1+x) \approx x$. Let's denote the expression as $y$, then: $$ \begin{align*} y &= (3.99)^3 (1.01)^4 (1.98)^{-1}\\ \ln(y) &= \ln((3.99)^3 (1.01)^4 (1.98)^{-1})\\ &= 3\ln(3.99) + 4\ln(1.01) - \ln(1.98) \end{align*} $$ Now, we can approximate each logarithm using the linear approximation: $$ \begin{align*} \ln(3.99) &\approx \ln(4) + \frac{1}{4}(3.99 - 4)\\ &= \ln(4) - 0.0025\\ \ln(1.01) &\approx \ln(1) + \frac{1}{1}(1.01 - 1)\\ &= 0.01\\ \ln(1.98) &\approx \ln(2) + \frac{1}{2}(1.98 - 2)\\ &= \ln(2) - 0.01 \end{align*} $$ Substituting these approximations back into the equation for $\ln(y)$: $$ \begin{align*} \ln(y) &\approx 3(\ln(4) - 0.0025) + 4(0.01) - (\ln(2) - 0.01)\\ &= 3\ln(4) - 0.0075 + 0.04 - \ln(2) + 0.01\\ &= 3\ln(4) - \ln(2) + 0.0425 \end{align*} $$ Now, we can exponentiate both sides to get an estimate for $y$: $$ \begin{align*} y &\approx e^{3\ln(4) - \ln(2) + 0.0425}\\ &= e^{\ln(4^3) - \ln(2) + 0.0425}\\ &= e^{\ln(64) - \ln(2) + 0.0425}\\ &= e^{\ln(32) + 0.0425}\\ &= 32 \cdot e^{0.0425}\\ &\approx 32 \cdot 1.0434\\ &\approx 33.389 \end{align*} $$ So, the linear approximation estimate for $(3.99)^3 (1.01)^4 (1.98)^{-1}$ is approximately $33.389$. |
33.36 |
| doubao | Let's consider the functions f(x) = x^3, g(x) = x^4, and h(x) = x^{-1}. The linear approximation of f(x) near x = 4 is f(4) + f'(4)(x - 4). f'(x) = 3x^2, so f'(4) = 3*4^2 = 48. f(4) = 4^3 = 64. So the linear approximation of f(x) near 4 is 64 + 48(x - 4). The linear approximation of g(x) near x = 1 is g(1) + g'(1)(x - 1). g'(x) = 4x^3, so g'(1) = 4. g(1) = 1. So the linear approximation of g(x) near 1 is 1 + 4(x - 1). The linear approximation of h(x) near x = 2 is h(2) + h'(2)(x - 2). h'(x) = -x^{-2}, so h'(2) = -1/4. h(2) = 1/2. So the linear approximation of h(x) near 2 is 1/2 - (1/4)(x - 2). Now we estimate: (3.99)^3 ≈ 64 + 48(3.99 - 4) = 64 - 48*0.01 = 64 - 0.48 = 63.52. (1.01)^4 ≈ 1 + 4(1.01 - 1) = 1 + 4*0.01 = 1.04. (1.98)^{-1} ≈ 1/2 - (1/4)(1.98 - 2) = 1/2 + (1/4)*0.02 = 1/2 + 0.005 = 0.505. Multiplying these approximations together gives: 63.52 * 1.04 * 0.505 ≈ 33.1. |
33.36 |
| minimax | To estimate the value of the expression \((3.99)^3 (1.01)^4 (1.98)^{-1}\) using linear approximation, we'll use the fact that for a function \(f(x)\) that is differentiable at \(x = a\), the linear approximation near \(x = a\) is given by: \[ f(x) \approx f(a) + f'(a)(x - a) \] We'll apply this approximation to each factor in the expression separately. 1. For \((3.99)^3\), let \(f(x) = x^3\) and \(a = 4\). The derivative of \(f(x)\) is \(f'(x) = 3x^2\). Using the linear approximation: \[ (3.99)^3 \approx 4^3 + 3(4)^2(-0.01) = 64 - 3(16)(0.01) = 64 - 0.48 = 63.52 \] 2. For \((1.01)^4\), let \(f(x) = x^4\) and \(a = 1\). The derivative of \(f(x)\) is \(f'(x) = 4x^3\). Using the linear approximation: \[ (1.01)^4 \approx 1^4 + 4(1)^3(0.01) = 1 + 4(0.01) = 1 + 0.04 = 1.04 \] 3. For \((1.98)^{-1}\), let \(f(x) = x^{-1}\) and \(a = 2\). The derivative of \(f(x)\) is \(f'(x) = -x^{-2}\). Using the linear approximation: \[ (1.98)^{-1} \approx 2^{-1} - 2^{-2}(-0.02) = 0.5 + 0.5(0.02) = 0.5 + 0.01 = 0.51 \] Now, we multiply the approximations together: \[ (3.99)^3 (1.01)^4 (1.98)^{-1} \approx 63.52 \times 1.04 \times 0.51 \] \[ \approx 63.52 \times 0.5304 \] \[ \approx 33.77 \] So, using linear approximation, we estimate that \((3.99)^3 (1.01)^4 (1.98)^{-1}\) is approximately 33.77. Keep in mind that this is an approximation and the actual value may differ slightly. |
33.36 |
| kimi | To estimate the value of \((3.99)^3 (1.01)^4 (1.98)^{-1}\) using linear approximation, we can use the fact that for small values of \(x\), the function \(f(x)\) can be approximated by its tangent line at a point \(a\), which is given by the formula: \[ f(x) \approx f(a) + f'(a)(x - a) \] For the exponential function \( f(x) = x^n \), the derivative is \( f'(x) = n \cdot x^{n-1} \). We can use this to approximate each term separately. Let's start by approximating each term: 1. For \((3.99)^3\), we can use the linear approximation around \(a = 4\) because \(3.99\) is close to \(4\): \[ (3.99)^3 \approx 4^3 + 3 \cdot 4^2 \cdot (3.99 - 4) \] \[ (3.99)^3 \approx 64 + 3 \cdot 16 \cdot (-0.01) \] \[ (3.99)^3 \approx 64 - 0.48 \] \[ (3.99)^3 \approx 63.52 \] 2. For \((1.01)^4\), we can use the linear approximation around \(a = 1\): \[ (1.01)^4 \approx 1^4 + 4 \cdot 1^3 \cdot (1.01 - 1) \] \[ (1.01)^4 \approx 1 + 4 \cdot 1 \cdot 0.01 \] \[ (1.01)^4 \approx 1 + 0.04 \] \[ (1.01)^4 \approx 1.04 \] 3. For \((1.98)^{-1}\), we can use the linear approximation around \(a = 2\) because \(1.98\) is close to \(2\): \[ (1.98)^{-1} \approx 2^{-1} + (-1) \cdot 2^{-2} \cdot (1.98 - 2) \] \[ (1.98)^{-1} \approx 0.5 - 0.25 \cdot (-0.02) \] \[ (1.98)^{-1} \approx 0.5 + 0.005 \] \[ (1.98)^{-1} \approx 0.505 \] Now, we multiply the approximations together: \[ (3.99)^3 (1.01)^4 (1.98)^{-1} \approx 63.52 \cdot 1.04 \cdot 0.505 \] Let's calculate the product: \[ 63.52 \cdot 1.04 \approx 66.0608 \] \[ 66.0608 \cdot 0.505 \approx 33.35 \] So, the linear approximation gives us an estimate of approximately \(33.35\) for \((3.99)^3 (1.01)^4 (1.98)^{-1}\). |
33.36 |
| wenxinyiyan | To estimate (3.99)^3 (1.01)^4 (1.98)^(-1) using linear approximation, we first need to choose a point close to these values where we can easily compute the function. Let's choose f(x, y, z) = x^3 * y^4 * z^(-1) and evaluate it at the point (4, 1, 2) which is close to (3.99, 1.01, 1.98). The linear approximation formula is: L(x, y, z) = f(a, b, c) + f_x(a, b, c)(x - a) + f_y(a, b, c)(y - b) + f_z(a, b, c)(z - c) where f_x, f_y, and f_z are the partial derivatives of f with respect to x, y, and z, respectively, and (a, b, c) is the point (4, 1, 2). First, let's compute the partial derivatives: f_x = 3x^2 * y^4 * z^(-1) f_y = 4x^3 * y^3 * z^(-1) f_z = -x^3 * y^4 * z^(-2) Now, let's evaluate these partial derivatives at (4, 1, 2): f_x(4, 1, 2) = 3(4^2) * (1^4) * (2^(-1)) = 24 f_y(4, 1, 2) = 4(4^3) * (1^3) * (2^(-1)) = 64 f_z(4, 1, 2) = -(4^3) * (1^4) * (2^(-2)) = -16 Also, compute f(4, 1, 2): f(4, 1, 2) = 4^3 * 1^4 * 2^(-1) = 32 Now, using the linear approximation formula: L(3.99, 1.01, 1.98) = 32 + 24(3.99 - 4) + 64(1.01 - 1) - 16(1.98 - 2) = 32 - 0.24 + 0.64 + 0.32 = 32.72 So, the linear approximation of (3.99)^3 (1.01)^4 (1.98)^(-1) is 32.72. |
33.36 |
| zhipu | To estimate the value of \( (3.99)^3 (1.01)^4 (1.98)^{-1} \) using linear approximation, we can use the fact that the linear approximation of a function \( f(x) \) at a point \( x = a \) is given by: \[ f(x) \approx f(a) + f'(a)(x - a) \] We will apply this approximation to each term separately and then combine the results. ### Step 1: Approximate \( (3.99)^3 \) Let \( f(x) = x^3 \). We will approximate \( f(3.99) \) using \( f(4) \) because 3.99 is close to 4. - \( f(4) = 4^3 = 64 \) - \( f'(x) = 3x^2 \), so \( f'(4) = 3 \cdot 4^2 = 48 \) Using linear approximation: \[ f(3.99) \approx f(4) + f'(4)(3.99 - 4) \] \[ f(3.99) \approx 64 + 48(-0.01) \] \[ f(3.99) \approx 64 - 0.48 \] \[ f(3.99) \approx 63.52 \] ### Step 2: Approximate \( (1.01)^4 \) Let \( g(x) = x^4 \). We will approximate \( g(1.01) \) using \( g(1) \) because 1.01 is close to 1. - \( g(1) = 1^4 = 1 \) - \( g'(x) = 4x^3 \), so \( g'(1) = 4 \cdot 1^3 = 4 \) Using linear approximation: \[ g(1.01) \approx g(1) + g'(1)(1.01 - 1) \] \[ g(1.01) \approx 1 + 4(0.01) \] \[ g(1.01) \approx 1 + 0.04 \] \[ g(1.01) \approx 1.04 \] ### Step 3: Approximate \( (1.98)^{-1} \) Let \( h(x) = x^{-1} \). We will approximate \( h(1.98) \) using \( h(2) \) because 1.98 is close to 2. - \( h(2) = 2^{-1} = 0.5 \) - \( h'(x) = -x^{-2} \), so \( h'(2) = -2^{-2} = -0.25 \) Using linear approximation: \[ h(1.98) \approx h(2) + h'(2)(1.98 - 2) \] \[ h(1.98) \approx 0.5 - 0.25(-0.02) \] \[ h(1.98) \approx 0.5 + 0.005 \] \[ h(1.98) \approx 0.505 \] ### Step 4: Combine the approximations Now, we combine the approximations: \[ (3.99)^3 (1.01)^4 (1.98)^{-1} \approx 63.52 \cdot 1.04 \cdot 0.505 \] Let's calculate this product: \[ 63.52 \cdot 1.04 \approx 66.0588 \] \[ 66.0588 \cdot 0.505 \approx 33.3399 \] So, the estimated value using linear approximation is approximately: \[ \boxed{33.34} \] This is a rough estimate, and the actual value might differ slightly, but it provides a good approximation using the linear method. |
33.36 |
| gpto1 | To estimate the value of \((3.99)^3 (1.01)^4 (1.98)^{-1}\) using linear approximation, we'll employ logarithms and properties of exponentials for small deviations. **Step 1: Recognize Small Deviations** Notice that: - \(3.99\) is close to \(4\) - \(1.01\) is close to \(1\) - \(1.98\) is close to \(2\) Let: - \(h_1 = 3.99 - 4 = -0.01\) - \(h_2 = 1.01 - 1 = 0.01\) - \(h_3 = 1.98 - 2 = -0.02\) **Step 2: Take Natural Logarithms** Consider the natural logarithm of the expression: \[ \ln\left[(3.99)^3 (1.01)^4 (1.98)^{-1}\right] = 3\ln(3.99) + 4\ln(1.01) - \ln(1.98) \] **Step 3: Linearize Each Logarithm** For small \(h\), the linear approximation is \(\ln(a + h) \approx \ln(a) + \dfrac{h}{a}\). Compute each term: 1. **\(\ln(3.99)\):** \[ \ln(3.99) = \ln(4 - 0.01) \approx \ln(4) - \dfrac{0.01}{4} = \ln(4) - 0.0025 \] 2. **\(\ln(1.01)\):** \[ \ln(1.01) = \ln(1 + 0.01) \approx \ln(1) + 0.01 = 0 + 0.01 \] 3. **\(\ln(1.98)\):** \[ \ln(1.98) = \ln(2 - 0.02) \approx \ln(2) - \dfrac{0.02}{2} = \ln(2) - 0.01 \] **Step 4: Combine the Terms** Substitute back into the original expression: \[ \begin{align*} 3\ln(3.99) + 4\ln(1.01) - \ln(1.98) &\approx 3\left(\ln(4) - 0.0025\right) + 4(0.01) - \left(\ln(2) - 0.01\right) \\ &= 3\ln(4) - 0.0075 + 0.04 - \ln(2) + 0.01. \end{align*} \] Simplify using \(\ln(4) = 2\ln(2)\): \[ \begin{align*} 3\ln(4) - \ln(2) &= 3(2\ln(2)) - \ln(2) = (6\ln(2)) - \ln(2) = 5\ln(2). \end{align*} \] Combine constants: \[ -0.0075 + 0.04 + 0.01 = 0.0425. \] So the total logarithm is: \[ 5\ln(2) + 0.0425. \] **Step 5: Exponentiate to Find the Approximate Value** \[ (3.99)^3 (1.01)^4 (1.98)^{-1} \approx e^{5\ln(2) + 0.0425} = 2^5 e^{0.0425} = 32 \times e^{0.0425}. \] **Step 6: Approximate \(e^{0.0425}\)** For small \(x\), \(e^{x} \approx 1 + x + \dfrac{x^2}{2}\): \[ e^{0.0425} \approx 1 + 0.0425 + \dfrac{(0.0425)^2}{2} = 1 + 0.0425 + 0.000903125 = 1.043403125. \] **Step 7: Calculate the Final Approximation** Multiply: \[ 32 \times 1.043403125 \approx 33.3889. \] **Conclusion:** The estimated value of the expression is approximately \(33.39\). **Answer:** Approximately **33.39** |
33.36 |
| deepseekv3 | To estimate the expression \((3.99)^3 \cdot (1.01)^4 \cdot (1.98)^{-1}\) using linear approximation, we will proceed step by step. Here's the detailed solution: --- ### Step 1: Rewrite the expression using nearby values We rewrite the expression in terms of small deviations from nearby values that are easier to compute: \[ (3.99)^3 \cdot (1.01)^4 \cdot (1.98)^{-1} = (4 - 0.01)^3 \cdot (1 + 0.01)^4 \cdot (2 - 0.02)^{-1} \] Here, the small deviations are: \[ \Delta x = -0.01, \quad \Delta y = 0.01, \quad \Delta z = -0.02 \] --- ### Step 2: Linearize each term We use the linear approximation formula \(f(a + \Delta) \approx f(a) + f'(a) \cdot \Delta\) for each term. #### 1. \((4 - 0.01)^3 = (4 + \Delta x)^3\) Let \(f(x) = x^3\). Then: \[ f'(x) = 3x^2 \] At \(x = 4\): \[ f(4) = 4^3 = 64, \quad f'(4) = 3 \cdot 4^2 = 48 \] Thus: \[ (4 - 0.01)^3 \approx 64 + 48 \cdot (-0.01) = 64 - 0.48 = 63.52 \] #### 2. \((1 + 0.01)^4 = (1 + \Delta y)^4\) Let \(f(y) = y^4\). Then: \[ f'(y) = 4y^3 \] At \(y = 1\): \[ f(1) = 1^4 = 1, \quad f'(1) = 4 \cdot 1^3 = 4 \] Thus: \[ (1 + 0.01)^4 \approx 1 + 4 \cdot 0.01 = 1 + 0.04 = 1.04 \] #### 3. \((2 - 0.02)^{-1} = (2 + \Delta z)^{-1}\) Let \(f(z) = z^{-1}\). Then: \[ f'(z) = -\frac{1}{z^2} \] At \(z = 2\): \[ f(2) = 2^{-1} = 0.5, \quad f'(2) = -\frac{1}{2^2} = -0.25 \] Thus: \[ (2 - 0.02)^{-1} \approx 0.5 + (-0.25) \cdot (-0.02) = 0.5 + 0.005 = 0.505 \] --- ### Step 3: Multiply the approximations Now multiply the linearized approximations: \[ 63.52 \cdot 1.04 \cdot 0.505 \] First, compute \(63.52 \cdot 1.04\): \[ 63.52 \cdot 1.04 = 63.52 \cdot 1 + 63.52 \cdot 0.04 = 63.52 + 2.5408 = 66.0608 \] Next, compute \(66.0608 \cdot 0.505\): \[ 66.0608 \cdot 0.505 = 66.0608 \cdot 0.5 + 66.0608 \cdot 0.005 = 33.0304 + 0.330304 = 33.360704 \] --- ### Step 4: Final approximation The estimated value of \((3.99)^3 \cdot (1.01)^4 \cdot (1.98)^{-1}\) is: \[ \boxed{33.36} \] --- This is the result of the linear approximation. |
33.36 |