| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To find \(g'(1)\), where \(g(x)\) is the inverse of \(f(x) = x + \cos(x)\), we can use the formula for the derivative of the inverse function. If \(g(x)\) is the inverse of \(f(x)\), then \(f(g(x)) = x\) for all \(x\) in the domain of \(g\), and we have: \[ f(g(x)) = x \] Differentiating both sides with respect to \(x\) using the chain rule, we get: \[ f'(g(x))g'(x) = 1 \] Therefore: \[ g'(x) = \frac{1}{f'(g(x))} \] Given \(f(x) = x + \cos(x)\), we find \(f'(x)\): \[ f'(x) = 1 - \sin(x) \] We are looking for \(g'(1)\). This requires us to find \(g(1)\) and then evaluate \(f'(g(1))\). Since \(g(x)\) is the inverse of \(f(x)\), we need to find \(x\) such that: \[ f(x) = 1 \] \[ x + \cos(x) = 1 \] Let's solve this equation for \(x\). This is not a straightforward algebraic equation, and it might require numerical methods to solve. However, for the purpose of this question, we can assume that such an \(x\) exists and call it \(x_0\). Thus, \(g(1) = x_0\), and we need to find \(f'(x_0)\): \[ f'(x_0) = 1 - \sin(x_0) \] Then: \[ g'(1) = \frac{1}{f'(g(1))} = \frac{1}{1 - \sin(x_0)} \] To find the exact value, we would need to solve \(x + \cos(x) = 1\) for \(x\), which is not trivial and usually requires numerical methods. However, the process outlined here provides a general approach to finding the derivative of the inverse function at a particular point. |
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| deepseekr1 | To find \( g'(1) \) where \( g(x) \) is the inverse of \( f(x) = x + \cos(x) \): 1. **Compute the derivative of \( f(x) \)**: \[ f'(x) = 1 - \sin(x) \] 2. **Determine \( g(1) \)**: Since \( g \) is the inverse of \( f \), we need \( x \) such that \( f(x) = 1 \): \[ x + \cos(x) = 1 \] Testing \( x = 0 \): \[ 0 + \cos(0) = 0 + 1 = 1 \] Thus, \( g(1) = 0 \). 3. **Evaluate \( f'(g(1)) \)**: Substitute \( g(1) = 0 \) into \( f'(x) \): \[ f'(0) = 1 - \sin(0) = 1 \] 4. **Apply the inverse function theorem**: \[ g'(1) = \frac{1}{f'(g(1))} = \frac{1}{f'(0)} = \frac{1}{1} = 1 \] Thus, the value of \( g'(1) \) is \(\boxed{1}\). |
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| gpto1 | To solve this problem, let's use the property of inverse functions and differentiation. **Step 1: Recognize the Inverse Relationship** Since \( g(x) \) is the inverse of \( f(x) = x + \cos(x) \), we have: \[ f(g(x)) = x \quad \text{and} \quad g(f(x)) = x \] **Step 2: Use the Derivative of an Inverse Function** The derivative of an inverse function is given by: \[ g'(x) = \frac{1}{f'(g(x))} \] So, to find \( g'(1) \), we need to find \( f'(g(1)) \). **Step 3: Find \( g(1) \)** Since \( f(g(1)) = 1 \), we need to solve for \( y = g(1) \) such that: \[ f(y) = y + \cos(y) = 1 \] Let's test \( y = 0 \): \[ 0 + \cos(0) = 0 + 1 = 1 \] This satisfies the equation, so \( y = g(1) = 0 \). **Step 4: Compute \( f'(y) \) at \( y = 0 \)** First, find the derivative \( f'(x) \): \[ f'(x) = 1 - \sin(x) \] Now, compute \( f'(0) \): \[ f'(0) = 1 - \sin(0) = 1 - 0 = 1 \] **Step 5: Find \( g'(1) \)** Using the derivative formula: \[ g'(1) = \frac{1}{f'(g(1))} = \frac{1}{f'(0)} = \frac{1}{1} = 1 \] **Answer:** \[ g\,'(1) \;=\; 1 \] |
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| GPT-4o | To find \( g'(1) \) where \( g(x) \) is the inverse function of \( f(x) = x + \cos(x) \), we'll use the formula for the derivative of an inverse function. Specifically, if \( g \) is the inverse of \( f \), then: \[ g'(y) = \frac{1}{f'(g(y))} \] We need to find \( g'(1) \), so we'll follow these steps: 1. Identify \( y \) such that \( f(y) = 1 \). 2. Compute \( f'(x) \). 3. Evaluate \( f'(y) \) where \( f(y) = 1 \). 4. Use the formula \( g'(1) = \frac{1}{f'(y)} \). ### Step 1: Find \( y \) such that \( f(y) = 1 \) We need to solve the equation \( f(y) = y + \cos(y) = 1 \): \[ y + \cos(y) = 1 \] This equation is transcendental and may not have a simple closed-form solution. However, we can find an approximate solution using numerical methods or inspection. By inspection, let's test \( y = 0 \): \[ 0 + \cos(0) = 0 + 1 = 1 \] So, \( y = 0 \) is a solution to \( y + \cos(y) = 1 \). ### Step 2: Compute \( f'(x) \) \[ f(x) = x + \cos(x) \] The derivative \( f'(x) \) is: \[ f'(x) = 1 - \sin(x) \] ### Step 3: Evaluate \( f'(y) \) at \( y = 0 \) \[ f'(0) = 1 - \sin(0) = 1 - 0 = 1 \] ### Step 4: Use the formula \( g'(1) = \frac{1}{f'(y)} \) Since \( y = 0 \) and \( f'(0) = 1 \): \[ g'(1) = \frac{1}{f'(0)} = \frac{1}{1} = 1 \] ### Conclusion \[ g'(1) = 1 \] So, the derivative of the inverse function \( g \) at \( x = 1 \) is: \[ g'(1) = 1 \] |
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| llama3.1-70B-Instruct | To find the derivative of the inverse function g(x), we can use the formula: g'(x) = 1 / f'(g(x)) First, we need to find the derivative of f(x) = x + cos(x): f'(x) = 1 - sin(x) Now, we need to find the value of g(1). Since g(x) is the inverse of f(x), we know that f(g(1)) = 1. Substituting f(x) = x + cos(x), we get: g(1) + cos(g(1)) = 1 Solving for g(1), we get: g(1) = 1 - cos(g(1)) Now, we can substitute g(1) into the formula for g'(x): g'(1) = 1 / f'(g(1)) = 1 / (1 - sin(g(1))) = 1 / (1 - sin(1 - cos(g(1)))) Simplifying, we get: g'(1) = 1 / (1 - sin(1 - cos(1 - cos(g(1))))) This is the final answer. |
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| llama3.1-405B-Instruct | To find the derivative of the inverse function g(x), we can use the formula: g'(x) = 1 / f'(g(x)) First, we need to find the derivative of f(x) = x + cos(x): f'(x) = 1 - sin(x) Now, we need to find the value of g(1). Since g(x) is the inverse of f(x), we know that f(g(1)) = 1. So, we need to solve the equation: g(1) + cos(g(1)) = 1 This equation cannot be solved analytically, so we will use numerical methods to find an approximate value of g(1). Using a numerical solver, we find that g(1) ≈ 0.739085. Now, we can find the value of f'(g(1)): f'(g(1)) = 1 - sin(g(1)) ≈ 1 - sin(0.739085) ≈ 0.673612 Finally, we can find the value of g'(1): g'(1) = 1 / f'(g(1)) ≈ 1 / 0.673612 ≈ 1.48329 Therefore, the value of g'(1) is approximately 1.48329. |
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| DeepSeek-Coder-V2 | To find \( g'(1) \) where \( g(x) \) is the inverse of \( f(x) = x + \cos(x) \), we can use the formula for the derivative of the inverse function: \[ g'(x) = \frac{1}{f'(g(x))} \] First, we need to find \( f'(x) \): \[ f(x) = x + \cos(x) \] \[ f'(x) = 1 - \sin(x) \] Next, we need to determine \( g(1) \). Since \( g(x) \) is the inverse of \( f(x) \), we have: \[ f(g(1)) = 1 \] We need to solve for \( g(1) \) such that: \[ g(1) + \cos(g(1)) = 1 \] This equation can be solved numerically or graphically. For simplicity, let's assume we have found \( g(1) \) to be approximately 0.514 (this value can be obtained through numerical methods). Now, we substitute \( g(1) \) into the derivative formula: \[ g'(1) = \frac{1}{f'(g(1))} \] \[ g'(1) = \frac{1}{1 - \sin(g(1))} \] Using the approximate value \( g(1) \approx 0.514 \): \[ \sin(0.514) \approx 0.491 \] \[ g'(1) = \frac{1}{1 - 0.491} \] \[ g'(1) = \frac{1}{0.509} \] \[ g'(1) \approx 1.964 \] Thus, the final answer is: \[ \boxed{1.964} \] |
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| Qwen2-72B-Instruct | To find \(g'(1)\), where \(g(x)\) is the inverse of \(f(x) = x + \cos(x)\), we can use the formula for the derivative of the inverse function. If \(g(x)\) is the inverse of \(f(x)\), then \(f(g(x)) = x\) for all \(x\) in the domain of \(g\), and we have: \[ f(g(x)) = x \] Differentiating both sides with respect to \(x\) using the chain rule, we get: \[ f'(g(x))g'(x) = 1 \] Therefore: \[ g'(x) = \frac{1}{f'(g(x))} \] Given \(f(x) = x + \cos(x)\), we find \(f'(x)\): \[ f'(x) = 1 - \sin(x) \] We are looking for \(g'(1)\). This requires us to find \(g(1)\) and then evaluate \(f'(g(1))\). Since \(g(x)\) is the inverse of \(f(x)\), we need to find \(x\) such that: \[ f(x) = 1 \] \[ x + \cos(x) = 1 \] Let's solve this equation for \(x\). This is not a straightforward algebraic equation, and it might require numerical methods to solve. However, for the purpose of this question, we can assume that such an \(x\) exists and call it \(x_0\). Thus, \(g(1) = x_0\), and we need to find \(f'(x_0)\): \[ f'(x_0) = 1 - \sin(x_0) \] Then: \[ g'(1) = \frac{1}{f'(g(1))} = \frac{1}{1 - \sin(x_0)} \] To find the exact value, we would need to solve \(x + \cos(x) = 1\) for \(x\), which is not trivial and usually requires numerical methods. However, the process outlined here provides a general approach to finding the derivative of the inverse function at a particular point. |
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| baichuan4 | To find the derivative of the inverse function \(g(x)\) at \(x=1\), we first need to find the derivative of the original function \(f(x) = x + \cos(x)\). The derivative of \(f(x)\) is: \[f'(x) = 1 - \sin(x)\] Now, since \(g(x)\) is the inverse of \(f(x)\), we have \(f(g(x)) = x\). Taking the derivative of both sides with respect to \(x\), we get: \[f'(g(x)) \cdot g'(x) = 1\] Rearranging to solve for \(g'(x)\), we have: \[g'(x) = \frac{1}{f'(g(x))}\] We want to find \(g'(1)\), so we need to find \(f'(g(1))\). But first, we need to determine what \(g(1)\) is. Since \(g(x)\) is the inverse of \(f(x)\), we set \(f(x) = 1\) and solve for \(x\): \[1 = x + \cos(x)\] \[x = 1 - \cos(x)\] This equation cannot be solved directly for \(x\) using simple algebra, as it is a transcendental equation. However, we don't actually need to solve it to find \(g'(1)\). Instead, we can proceed by substituting \(x=1\) into the derivative of \(f(x)\) to find \(f'(g(1))\): \[f'(g(1)) = 1 - \sin(g(1))\] Since we don't know the exact value of \(g(1)\), we cannot directly calculate this. However, we can use the fact that \(f(g(1)) = 1\) to get an expression for \(g'(1)\): \[g'(1) = \frac{1}{f'(g(1))} = \frac{1}{1 - \sin(g(1))}\] At this point, without the exact value of \(g(1)\), we cannot provide a numerical answer for \(g'(1)\). To complete the solution, we would need to find \(g(1)\) numerically or analytically, which would involve solving the equation \(1 = x + \cos(x)\). This equation would typically be solved using numerical methods such as Newton's method. However, since we are not able to perform numerical calculations here, we cannot give a specific numerical value for \(g'(1)\). In practice, you would use a calculator or mathematical software to find \(g(1)\), then plug that value into the expression for \(g'(1)\) above to obtain the final answer. |
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| doubao | First, we need to find the value of x for which f(x) = 1. That is, x + cos(x) = 1. This equation may not have an easy analytic solution, but we can try to find an approximate value numerically. Once we find the value of x (let's call it x0) such that f(x0) = 1, then g'(1) = 1 / f'(x0). To find f'(x), we take the derivative of f(x) = x + cos(x), which is f'(x) = 1 - sin(x). So, we need to find x0 such that x0 + cos(x0) = 1 and then calculate g'(1) = 1 / (1 - sin(x0)). This is a rather involved process and may require numerical methods or approximation techniques to find x0 and then evaluate g'(1) accurately. |
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| minimax | To find the derivative of the inverse function \( g(x) \) at a point, we can use the formula that relates the derivative of a function and its inverse: \[ (f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))} \] First, we need to find \( f'(x) \), the derivative of the original function \( f(x) = x + \cos(x) \). \[ f'(x) = \frac{d}{dx}(x + \cos(x)) \] \[ f'(x) = 1 - \sin(x) \] Now, we need to evaluate \( f'(x) \) at the point where \( x = g(1) \), because we want to find \( g'(1) \), and by the formula above, we need \( f'(g(1)) \). Since \( g(x) \) is the inverse of \( f(x) \), by definition \( f(g(x)) = x \). Therefore, \( f(g(1)) = 1 \). We need to find the value of \( x \) such that \( f(x) = 1 \). \[ f(x) = x + \cos(x) = 1 \] \[ \cos(x) = 1 - x \] This equation does not have a simple algebraic solution, and we would typically need to solve it numerically to find the value of \( x \) such that \( f(x) = 1 \). However, since we are looking for \( g'(1) \), we can use the fact that \( g(1) \) is the value of \( x \) that satisfies \( f(x) = 1 \). Let's assume we have found \( g(1) = x_0 \) such that \( f(x_0) = 1 \). Then we can evaluate \( f'(x_0) \): \[ f'(x_0) = 1 - \sin(x_0) \] Now we can use the formula for the derivative of the inverse function: \[ g'(1) = \frac{1}{f'(g(1))} = \frac{1}{1 - \sin(x_0)} \] Without a numerical method or a calculator, we cannot find the exact value of \( x_0 \) and thus cannot provide a numerical value for \( g'(1) \). However, if you have access to a numerical solver or a calculator, you can find \( x_0 \) such that \( f(x_0) = 1 \), compute \( \sin(x_0) \), and then find \( g'(1) \) using the formula above. |
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| kimi | To find the derivative of the inverse function \( g(x) \) of \( f(x) = x + \cos(x) \) at \( g'(1) \), we can use the formula for the derivative of the inverse function, which states that if \( f \) is invertible and \( f'(a) \neq 0 \), then \( g'(a) = \frac{1}{f'(f^{-1}(a))} \). Here are the steps to find \( g'(1) \): 1. **Find \( f'(x) \)**: The derivative of \( f(x) = x + \cos(x) \) is \( f'(x) = 1 - \sin(x) \). 2. **Find \( f^{-1}(1) \)**: We need to find the value of \( x \) such that \( f(x) = 1 \). This is equivalent to solving \( x + \cos(x) = 1 \). This equation might not have a simple algebraic solution, but we can use numerical methods to approximate it if needed. 3. **Evaluate \( f'(f^{-1}(1)) \)**: Once we have \( f^{-1}(1) \), we can plug it into \( f'(x) \) to get \( f'(f^{-1}(1)) \). 4. **Calculate \( g'(1) \)**: Finally, we use the formula \( g'(1) = \frac{1}{f'(f^{-1}(1))} \). Let's go through these steps: 1. \( f'(x) = 1 - \sin(x) \). 2. To find \( f^{-1}(1) \), we need to solve \( x + \cos(x) = 1 \). This is not straightforward to solve algebraically, so we might use numerical methods to approximate the solution. 3. Once we have \( f^{-1}(1) \), let's call it \( a \), we can find \( f'(a) \) by plugging \( a \) into \( f'(x) \). 4. \( g'(1) = \frac{1}{f'(a)} \). However, without a specific value for \( f^{-1}(1) \), we cannot calculate \( g'(1) \) exactly. If you have a specific value for \( f^{-1}(1) \), you can plug it into the formula to find \( g'(1) \). Otherwise, you would need to use numerical methods to approximate \( f^{-1}(1) \) and then calculate \( g'(1) \). |
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| wenxinyiyan | 首先，我们设y = g(x)，这意味着x = f(y) = y + cos(y)。 为了找到g'(x)，我们需要对x = y + cos(y)两边关于y求导。 对x关于y求导，我们得到： dx/dy = 1 - sin(y) 但是，我们需要找到g'(x)，即dy/dx。由于dx/dy = 1/(dy/dx)，我们可以得到： dy/dx = 1/(1 - sin(y)) 现在，为了找到g'(1)，我们需要找到当x = 1时y的值。由于g(x)是f(x)的逆函数，这意味着当f(y) = 1时，y = g(1)。 因此，我们需要解方程： 1 = y + cos(y) 这个方程没有简单的解析解，但我们可以使用数值方法找到y的近似值。然而，为了简化问题，我们可以注意到cos(y)的最大值是1，所以当y = 0时，1 = y + cos(y)。 因此，当x = 1时，y = 0。 现在我们可以找到g'(1)： g'(1) = 1/(1 - sin(0)) = 1/(1 - 0) = 1 所以，答案是g'(1) = 1。 |
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| zhipu | To find \( g'(1) \), where \( g(x) \) is the inverse of \( f(x) = x + \cos(x) \), we can use the relationship between a function and its inverse regarding derivatives. If \( g(x) \) is the inverse of \( f(x) \), then: \[ g'(x) = \frac{1}{f'(g(x))} \] To find \( g'(1) \), we need to find the value of \( x \) such that \( f(x) = 1 \), and then evaluate \( f'(x) \) at that value. ### Step 1: Find \( x \) such that \( f(x) = 1 \) \[ f(x) = x + \cos(x) = 1 \] We need to solve for \( x \): \[ x + \cos(x) = 1 \] This equation may not have a simple algebraic solution, but we can find the value numerically or by inspection. For simplicity, let's assume we find that \( x = a \) satisfies this equation, i.e., \[ a + \cos(a) = 1 \] ### Step 2: Find \( f'(x) \) The derivative of \( f(x) = x + \cos(x) \) is: \[ f'(x) = 1 - \sin(x) \] ### Step 3: Evaluate \( f'(x) \) at \( x = a \) \[ f'(a) = 1 - \sin(a) \] ### Step 4: Use the inverse function theorem \[ g'(1) = \frac{1}{f'(a)} \] Since \( a \) is the value such that \( f(a) = 1 \), we substitute \( f'(a) \): \[ g'(1) = \frac{1}{1 - \sin(a)} \] ### Summary To find the exact value of \( g'(1) \), we need the specific value of \( a \) that satisfies \( a + \cos(a) = 1 \). However, the general form of the solution is: \[ g'(1) = \frac{1}{1 - \sin(a)} \] where \( a \) is the solution to \( a + \cos(a) = 1 \). If you have a specific numerical method or a calculator to find \( a \), you can substitute it back into the expression to get a numerical value for \( g'(1) \). For example, if \( a \approx 0.567 \) (this is a hypothetical value for illustration), then: \[ \sin(0.567) \approx 0.537 \] \[ g'(1) \approx \frac{1}{1 - 0.537} \approx \frac{1}{0.463} \approx 2.16 \] This is a hypothetical numerical example. The exact value would depend on the precise solution to \( a + \cos(a) = 1 \). |
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| gpto1 | To solve this problem, let's use the property of inverse functions and differentiation. **Step 1: Recognize the Inverse Relationship** Since \( g(x) \) is the inverse of \( f(x) = x + \cos(x) \), we have: \[ f(g(x)) = x \quad \text{and} \quad g(f(x)) = x \] **Step 2: Use the Derivative of an Inverse Function** The derivative of an inverse function is given by: \[ g'(x) = \frac{1}{f'(g(x))} \] So, to find \( g'(1) \), we need to find \( f'(g(1)) \). **Step 3: Find \( g(1) \)** Since \( f(g(1)) = 1 \), we need to solve for \( y = g(1) \) such that: \[ f(y) = y + \cos(y) = 1 \] Let's test \( y = 0 \): \[ 0 + \cos(0) = 0 + 1 = 1 \] This satisfies the equation, so \( y = g(1) = 0 \). **Step 4: Compute \( f'(y) \) at \( y = 0 \)** First, find the derivative \( f'(x) \): \[ f'(x) = 1 - \sin(x) \] Now, compute \( f'(0) \): \[ f'(0) = 1 - \sin(0) = 1 - 0 = 1 \] **Step 5: Find \( g'(1) \)** Using the derivative formula: \[ g'(1) = \frac{1}{f'(g(1))} = \frac{1}{f'(0)} = \frac{1}{1} = 1 \] **Answer:** \[ g\,'(1) \;=\; 1 \] |
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| deepseekv3 | Alright, I have this problem: *Let \( g(x) \) be the inverse of \( f(x) = x + \cos(x) \). What is \( g'(1) \)?* At first glance, it seems like I need to find the derivative of the inverse function \( g(x) \) at \( x = 1 \). I remember that there's a formula for the derivative of an inverse function, but I'm not entirely sure how to apply it here. Let's break it down step by step. ### Understanding the Problem First, I need to understand what's being asked. I have a function \( f(x) = x + \cos(x) \), and \( g(x) \) is its inverse. So, by definition: \[ f(g(x)) = x \] \[ g(f(x)) = x \] I need to find \( g'(1) \), which is the derivative of \( g(x) \) evaluated at \( x = 1 \). ### Recalling the Formula for the Derivative of an Inverse Function I remember that if \( g(x) \) is the inverse of \( f(x) \), then the derivative of \( g(x) \) at a point \( x = a \) is given by: \[ g'(a) = \frac{1}{f'(g(a))} \] So, to find \( g'(1) \), I need to find \( g(1) \) and then evaluate \( f'(x) \) at that point. ### Finding \( g(1) \) First, I need to find \( g(1) \). Since \( g(x) \) is the inverse of \( f(x) \), \( g(1) \) is the value of \( x \) such that \( f(x) = 1 \). So, I need to solve: \[ x + \cos(x) = 1 \] This equation seems a bit tricky because it involves both \( x \) and \( \cos(x) \). I'm not sure how to solve it algebraically, so maybe I can find an approximate value. Let's try to find \( x \) such that \( x + \cos(x) = 1 \). I'll start by testing some values: 1. **When \( x = 0 \):** \[ 0 + \cos(0) = 0 + 1 = 1 \] Wait, that's exactly \( 1 \). So, \( x = 0 \) is a solution. But before I conclude, I should check if there are other solutions. The function \( f(x) = x + \cos(x) \) is increasing because its derivative \( f'(x) = 1 - \sin(x) \) is always non-negative (since \( \sin(x) \leq 1 \)), and it's only zero at specific points. So, \( f(x) \) is strictly increasing, which means it's one-to-one, and its inverse \( g(x) \) is well-defined. Therefore, the only solution to \( f(x) = 1 \) is \( x = 0 \). So, \( g(1) = 0 \). ### Calculating \( f'(x) \) Next, I need to find \( f'(x) \): \[ f(x) = x + \cos(x) \] \[ f'(x) = 1 - \sin(x) \] ### Applying the Inverse Function Derivative Formula Now, I can use the formula: \[ g'(1) = \frac{1}{f'(g(1))} \] I already found that \( g(1) = 0 \), so: \[ f'(g(1)) = f'(0) = 1 - \sin(0) = 1 - 0 = 1 \] Therefore: \[ g'(1) = \frac{1}{1} = 1 \] ### Double-Checking My Work Let me make sure I didn't make any mistakes. 1. **Finding \( g(1) \):** - I set \( f(x) = 1 \) and solved \( x + \cos(x) = 1 \). - \( x = 0 \) satisfies this equation. - Since \( f(x) \) is strictly increasing, \( x = 0 \) is the only solution. - So, \( g(1) = 0 \) is correct. 2. **Calculating \( f'(x) \):** - \( f(x) = x + \cos(x) \) - \( f'(x) = 1 - \sin(x) \) - This seems correct. 3. **Applying the formula:** - \( g'(1) = \frac{1}{f'(g(1))} = \frac{1}{f'(0)} \) - \( f'(0) = 1 - \sin(0) = 1 \) - So, \( g'(1) = 1 \) Everything checks out. ### Final Answer \[ \boxed{1} \] |
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