| 模型 | 模型答案 | 正确答案 |
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| 金声玉亮2.0 | To find the value of \(a > 0\) such that the tangent line to the graph of \(f(x) = x^2e^{-x}\) at \(x = a\) passes through the origin, we need to follow these steps: 1. Find the derivative of \(f(x)\) to determine the slope of the tangent line at any point \(x\). 2. Use the point-slope form of a line to write the equation of the tangent line at \(x = a\). 3. Set the \(y\)-intercept of the tangent line to 0 to find \(a\). Step 1: Find the derivative of \(f(x)\). The function is \(f(x) = x^2e^{-x}\). We can use the product rule to find its derivative: \[ f'(x) = 2xe^{-x} + x^2(-e^{-x}) = 2xe^{-x} - x^2e^{-x} \] Simplifying: \[ f'(x) = e^{-x}(2x - x^2) \] Step 2: Write the equation of the tangent line at \(x = a\). The slope of the tangent line at \(x = a\) is \(f'(a) = e^{-a}(2a - a^2)\). The point on the curve at \(x = a\) is \((a, f(a)) = (a, a^2e^{-a})\). The equation of the tangent line at \(x = a\) is: \[ y - a^2e^{-a} = e^{-a}(2a - a^2)(x - a) \] Step 3: Set the \(y\)-intercept of the tangent line to 0. When the line passes through the origin, the \(y\)-intercept is 0. Let’s substitute \(x = 0\) and \(y = 0\) into the equation of the tangent line: \[ 0 - a^2e^{-a} = e^{-a}(2a - a^2)(0 - a) \] Simplifying: \[ -a^2e^{-a} = -ae^{-a}(2a - a^2) \] Dividing both sides by \(-ae^{-a}\) (assuming \(a \neq 0\) and \(a \neq \infty\)): \[ a = 2a - a^2 \] Solving for \(a\): \[ a^2 - a = 0 \] \[ a(a - 1) = 0 \] This gives us \(a = 0\) and \(a = 1\). Since we’re looking for \(a > 0\), our solution is: \[ a = 1 \] |
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| deepseekr1 | To find the value of \( a > 0 \) such that the tangent line to the graph of \( f(x) = x^2 e^{-x} \) at \( x = a \) passes through the origin, we start by noting that the equation of the tangent line at \( x = a \) is given by: \[ y = f(a) + f'(a)(x - a) \] Since this tangent line passes through the origin (0, 0), substituting \( x = 0 \) and \( y = 0 \) into the equation gives: \[ 0 = f(a) - a f'(a) \] This simplifies to: \[ f(a) = a f'(a) \] Next, we compute \( f(a) \) and \( f'(a) \). The function is \( f(x) = x^2 e^{-x} \). Using the product rule, the derivative is: \[ f'(x) = 2x e^{-x} - x^2 e^{-x} = e^{-x} (2x - x^2) \] Substituting \( x = a \), we have: \[ f(a) = a^2 e^{-a} \] \[ f'(a) = e^{-a} (2a - a^2) \] Setting \( f(a) = a f'(a) \): \[ a^2 e^{-a} = a \left( e^{-a} (2a - a^2) \right) \] Dividing both sides by \( e^{-a} \): \[ a^2 = a (2a - a^2) \] Simplifying the right side: \[ a^2 = 2a^2 - a^3 \] Rearranging terms: \[ a^3 - a^2 = 0 \] \[ a^2 (a - 1) = 0 \] This gives \( a = 0 \) or \( a = 1 \). Since \( a > 0 \), the solution is \( a = 1 \). Verifying, the tangent line at \( a = 1 \) is \( y = \frac{x}{e} \), which passes through the origin. Thus, the value of \( a \) is \(\boxed{1}\). |
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| gpto1 | To find the value of \( a > 0 \) such that the tangent line to the graph of \( f(x) = x^2 e^{-x} \) at \( x = a \) passes through the origin \((0, 0)\), we proceed as follows. **Step 1: Find the equation of the tangent line at \( x = a \).** The equation of the tangent line to \( f(x) \) at \( x = a \) is given by: \[ y = f(a) + f'(a)(x - a) \] **Step 2: Set the tangent line to pass through the origin.** For the tangent line to pass through the origin \((0, 0)\): \[ 0 = f(a) + f'(a)(0 - a) \] Simplify: \[ 0 = f(a) - a f'(a) \] This leads to the equation: \[ a f'(a) = f(a) \] **Step 3: Compute \( f(a) \) and \( f'(a) \).** First, compute \( f(a) \): \[ f(a) = a^2 e^{-a} \] Next, compute \( f'(x) \) using the product rule: \[ f(x) = x^2 e^{-x} \] \[ f'(x) = \left(2x\right)e^{-x} + x^2(-e^{-x}) = e^{-x}\left(2x - x^2\right) \] So, \[ f'(a) = e^{-a}\left(2a - a^2\right) \] **Step 4: Substitute \( f(a) \) and \( f'(a) \) back into the equation.** Substitute \( f(a) \) and \( f'(a) \) into \( a f'(a) = f(a) \): \[ a \left[e^{-a}\left(2a - a^2\right)\right] = a^2 e^{-a} \] Simplify: \[ e^{-a} a \left(2a - a^2\right) = a^2 e^{-a} \] Subtract \( a^2 e^{-a} \) from both sides: \[ e^{-a} a \left(2a - a^2\right) - a^2 e^{-a} = 0 \] Factor out \( e^{-a} \): \[ e^{-a} \left[ a(2a - a^2) - a^2 \right] = 0 \] Simplify inside the brackets: \[ a(2a - a^2) - a^2 = 2a^2 - a^3 - a^2 = a^2 - a^3 \] So the equation becomes: \[ e^{-a} \left[ a^2 (1 - a) \right] = 0 \] **Step 5: Solve for \( a \).** Since \( e^{-a} > 0 \) for all real \( a \), we set: \[ a^2 (1 - a) = 0 \] This gives: \[ a^2 = 0 \quad \text{or} \quad 1 - a = 0 \] Solving these equations: 1. \( a^2 = 0 \) leads to \( a = 0 \). 2. \( 1 - a = 0 \) leads to \( a = 1 \). **Step 6: Consider the domain restriction.** Since \( a > 0 \), \( a = 0 \) is not acceptable. Therefore, the only valid solution is \( a = 1 \). **Answer:** \[ a = 1 \] |
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| GPT-4o | To find the value of \( a > 0 \) such that the tangent line to the graph of \( f(x) = x^2 e^{-x} \) at \( x = a \) passes through the origin, we need to follow these steps: 1. **Find the derivative \( f'(x) \) to get the slope of the tangent line at any point \( x \).** Given \( f(x) = x^2 e^{-x} \), we use the product rule to find the derivative. \[ f'(x) = \frac{d}{dx} (x^2 e^{-x}) = \frac{d}{dx} (x^2) \cdot e^{-x} + x^2 \cdot \frac{d}{dx} (e^{-x}) \] \[ f'(x) = 2x \cdot e^{-x} + x^2 \cdot (-e^{-x}) \] \[ f'(x) = 2x e^{-x} - x^2 e^{-x} \] \[ f'(x) = e^{-x} (2x - x^2) \] 2. **Find the equation of the tangent line at \( x = a \).** The slope of the tangent line at \( x = a \) is \( f'(a) \): \[ f'(a) = e^{-a} (2a - a^2) \] The point of tangency on the curve \( f(x) \) at \( x = a \) is \( (a, f(a)) \): \[ f(a) = a^2 e^{-a} \] The equation of the tangent line at \( x = a \) using the point-slope form \( y - y_1 = m(x - x_1) \) is: \[ y - f(a) = f'(a)(x - a) \] Substituting \( f(a) \) and \( f'(a) \): \[ y - a^2 e^{-a} = e^{-a} (2a - a^2)(x - a) \] 3. **Determine the condition for the tangent line to pass through the origin \((0,0)\).** Substitute \((0,0)\) into the tangent line equation: \[ 0 - a^2 e^{-a} = e^{-a} (2a - a^2)(0 - a) \] Simplify: \[ -a^2 e^{-a} = e^{-a} (2a - a^2)(-a) \] \[ -a^2 e^{-a} = -a e^{-a} (2a - a^2) \] Cancel out \( e^{-a} \) (since \( e^{-a} \neq 0 \)): \[ -a^2 = -a (2a - a^2) \] Simplify the right side: \[ -a^2 = -2a^2 + a^3 \] Combine like terms: \[ 0 = a^3 - a^2 \] Factor out \( a^2 \): \[ 0 = a^2 (a - 1) \] 4. **Solve for \( a \).** Since \( a^2 (a - 1) = 0 \), we get: \[ a^2 = 0 \quad \text{or} \quad a - 1 = 0 \] Since \( a > 0 \), we discard \( a = 0 \), leaving: \[ a = 1 \] Therefore, the value of \( a \) such that the tangent line to the graph of \( f(x) = x^2 e^{-x} \) at \( x = a \) passes through the origin is \( \boxed{1} \). |
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| llama3.1-70B-Instruct | To find the value of a > 0, such that the tangent line to the graph of f(x) = (x^2) (e^(-x)) at x = a passes through the origin, we need to follow these steps: 1. Find the derivative of f(x) with respect to x. 2. Evaluate the derivative at x = a to find the slope of the tangent line. 3. Use the point-slope form of a line to find the equation of the tangent line. 4. Set the equation of the tangent line equal to 0, since it passes through the origin. 5. Solve for a. Step 1: Find the derivative of f(x) with respect to x. Using the product rule, we get: f'(x) = d/dx [(x^2) (e^(-x))] = (x^2) d/dx (e^(-x)) + (e^(-x)) d/dx (x^2) = (x^2) (-e^(-x)) + (e^(-x)) (2x) = -x^2 e^(-x) + 2x e^(-x) Step 2: Evaluate the derivative at x = a to find the slope of the tangent line. f'(a) = -a^2 e^(-a) + 2a e^(-a) Step 3: Use the point-slope form of a line to find the equation of the tangent line. The point-slope form of a line is: y - y1 = m(x - x1) where (x1, y1) is a point on the line, and m is the slope. In this case, the point is (a, f(a)), and the slope is f'(a). So, the equation of the tangent line is: y - f(a) = f'(a) (x - a) Step 4: Set the equation of the tangent line equal to 0, since it passes through the origin. 0 - f(a) = f'(a) (0 - a) Simplifying, we get: -f(a) = -a f'(a) Step 5: Solve for a. Substituting the expressions for f(a) and f'(a), we get: -(a^2 e^(-a)) = -a (-a^2 e^(-a) + 2a e^(-a)) Simplifying, we get: a^2 e^(-a) = a^3 e^(-a) - 2a^2 e^(-a) Combine like terms: 2a^2 e^(-a) = a^3 e^(-a) Divide both sides by a^2 e^(-a): 2 = a So, the value of a > 0, such that the tangent line to the graph of f(x) = (x^2) (e^(-x)) at x = a passes through the origin, is a = 2. |
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| llama3.1-405B-Instruct | To find the value of a > 0 such that the tangent line to the graph of f(x) = (x^2) (e^(-x)) at x = a passes through the origin, we need to find the equation of the tangent line and then set it equal to 0. First, let's find the derivative of f(x) using the product rule: f'(x) = d/dx (x^2 e^(-x)) = 2x e^(-x) - x^2 e^(-x) Now, we need to find the slope of the tangent line at x = a. We can do this by plugging x = a into the derivative: f'(a) = 2a e^(-a) - a^2 e^(-a) The equation of the tangent line at x = a is given by: y - f(a) = f'(a) (x - a) We want the tangent line to pass through the origin, so we set x = 0 and y = 0: 0 - f(a) = f'(a) (0 - a) Simplifying, we get: f(a) = a f'(a) Now, we can plug in the expressions for f(a) and f'(a): a^2 e^(-a) = a (2a e^(-a) - a^2 e^(-a)) Simplifying, we get: a^2 e^(-a) = 2a^2 e^(-a) - a^3 e^(-a) Rearranging, we get: a^3 e^(-a) = a^2 e^(-a) Dividing both sides by a^2 e^(-a), we get: a = 1 Therefore, the value of a > 0 such that the tangent line to the graph of f(x) = (x^2) (e^(-x)) at x = a passes through the origin is a = 1. |
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| DeepSeek-Coder-V2 | To solve the problem, we need to find the value of \( a > 0 \) such that the tangent line to the graph of \( f(x) = x^2 e^{-x} \) at \( x = a \) passes through the origin. First, we need to find the equation of the tangent line to the graph of \( f(x) \) at \( x = a \). The general form of the tangent line equation at a point \( (a, f(a)) \) is: \[ y - f(a) = f'(a)(x - a) \] 1. **Calculate \( f(a) \):** \[ f(a) = a^2 e^{-a} \] 2. **Calculate \( f'(a) \):** To find the derivative \( f'(x) \), we use the product rule: \[ f(x) = x^2 e^{-x} \] \[ f'(x) = \frac{d}{dx}(x^2) e^{-x} + x^2 \frac{d}{dx}(e^{-x}) \] \[ f'(x) = 2x e^{-x} + x^2 (-e^{-x}) \] \[ f'(x) = 2x e^{-x} - x^2 e^{-x} \] \[ f'(x) = e^{-x} (2x - x^2) \] Therefore, \[ f'(a) = e^{-a} (2a - a^2) \] 3. **Equation of the tangent line at \( x = a \):** \[ y - a^2 e^{-a} = e^{-a} (2a - a^2) (x - a) \] 4. **Tangent line passes through the origin (0, 0):** Substitute \( (0, 0) \) into the tangent line equation: \[ 0 - a^2 e^{-a} = e^{-a} (2a - a^2) (0 - a) \] Simplify the equation: \[ -a^2 e^{-a} = e^{-a} (2a - a^2) (-a) \] \[ -a^2 e^{-a} = -a e^{-a} (2a - a^2) \] \[ -a^2 = -a (2a - a^2) \] \[ a^2 = a (2a - a^2) \] \[ a^2 = 2a^2 - a^3 \] \[ 0 = a^3 - a^2 \] \[ 0 = a^2 (a - 1) \] 5. **Solve for \( a \):** \[ a^2 (a - 1) = 0 \] This gives us two solutions: \[ a^2 = 0 \quad \text{or} \quad a - 1 = 0 \] \[ a = 0 \quad \text{or} \quad a = 1 \] Since we are given that \( a > 0 \), the only valid solution is: \[ a = 1 \] Therefore, the value of \( a \) such that the tangent line to the graph of \( f(x) = x^2 e^{-x} \) at \( x = a \) passes through the origin is: \[ \boxed{1} \] |
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| Qwen2-72B-Instruct | To find the value of \(a > 0\) such that the tangent line to the graph of \(f(x) = x^2e^{-x}\) at \(x = a\) passes through the origin, we need to follow these steps: 1. Find the derivative of \(f(x)\) to determine the slope of the tangent line at any point \(x\). 2. Use the point-slope form of a line to write the equation of the tangent line at \(x = a\). 3. Set the \(y\)-intercept of the tangent line to 0 to find \(a\). Step 1: Find the derivative of \(f(x)\). The function is \(f(x) = x^2e^{-x}\). We can use the product rule to find its derivative: \[ f'(x) = 2xe^{-x} + x^2(-e^{-x}) = 2xe^{-x} - x^2e^{-x} \] Simplifying: \[ f'(x) = e^{-x}(2x - x^2) \] Step 2: Write the equation of the tangent line at \(x = a\). The slope of the tangent line at \(x = a\) is \(f'(a) = e^{-a}(2a - a^2)\). The point on the curve at \(x = a\) is \((a, f(a)) = (a, a^2e^{-a})\). The equation of the tangent line at \(x = a\) is: \[ y - a^2e^{-a} = e^{-a}(2a - a^2)(x - a) \] Step 3: Set the \(y\)-intercept of the tangent line to 0. When the line passes through the origin, the \(y\)-intercept is 0. Let’s substitute \(x = 0\) and \(y = 0\) into the equation of the tangent line: \[ 0 - a^2e^{-a} = e^{-a}(2a - a^2)(0 - a) \] Simplifying: \[ -a^2e^{-a} = -ae^{-a}(2a - a^2) \] Dividing both sides by \(-ae^{-a}\) (assuming \(a \neq 0\) and \(a \neq \infty\)): \[ a = 2a - a^2 \] Solving for \(a\): \[ a^2 - a = 0 \] \[ a(a - 1) = 0 \] This gives us \(a = 0\) and \(a = 1\). Since we’re looking for \(a > 0\), our solution is: \[ a = 1 \] |
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| baichuan4 | To find the value of \(a > 0\) such that the tangent line to the graph of \(f(x) = x^2 e^{-x}\) at \(x = a\) passes through the origin, we need to do the following steps: 1. Find the derivative of \(f(x)\) to determine the slope of the tangent line at any point \(x\). 2. Substitute \(a\) for \(x\) in the derivative to find the slope of the tangent line at \(x = a\). 3. Find the equation of the tangent line at \(x = a\). 4. Use the fact that the tangent line passes through the origin (0,0) to solve for \(a\). Let's begin: **Step 1: Finding the derivative of \(f(x)\):** We use the product rule to differentiate \(f(x) = x^2 e^{-x}\): \[f'(x) = 2xe^{-x} + x^2(-e^{-x})\] \[f'(x) = 2xe^{-x} - x^2e^{-x}\] **Step 2: Substitute \(a\) for \(x\) in the derivative:** \[f'(a) = 2ae^{-a} - a^2e^{-a}\] **Step 3: Find the equation of the tangent line at \(x = a\):** The equation of a line with slope \(m\) through a point \((x_1, y_1)\) is given by: \[y - y_1 = m(x - x_1)\] The slope of the tangent line is \(f'(a)\) and it passes through the point \((a, f(a))\), so the equation of the tangent line is: \[y - a^2e^{-a} = (2ae^{-a} - a^2e^{-a})(x - a)\] **Step 4: Use the fact that the tangent line passes through the origin (0,0):** Substitute \((x, y) = (0, 0)\) into the tangent line equation: \[0 - a^2e^{-a} = (2ae^{-a} - a^2e^{-a})(0 - a)\] Simplify: \[-a^2e^{-a} = -a(2ae^{-a} - a^2e^{-a})\] \[-a^2e^{-a} = -2a^2e^{-a} + a^3e^{-a}\] Rearrange to solve for \(a\): \[a^3e^{-a} - a^2e^{-a} = -2a^2e^{-a} + a^3e^{-a}\] \[a^2e^{-a} = 2a^2e^{-a}\] Divide by \(a^2e^{-a}\) on both sides (note that \(a > 0\) so \(a^2e^{-a} > 0\), and thus we can divide without switching sides): \[1 = 2\] We have arrived at a contradiction, which means there was an error in our steps. Let's correct the mistake. The correct step 4 should not involve substituting the tangent line equation with the origin coordinates directly after finding the slope. Instead, we should have used the point-slope form of the tangent line equation at \(x = a\) and then used the fact that it passes through the origin to find \(a\). Here's the correction for step 4: The tangent line equation at \(x = a\) is: \[y = (2ae^{-a} - a^2e^{-a})(x - a) + a^2e^{-a}\] Since this line passes through the origin, we substitute \(y = 0\) and \(x = 0\): \[0 = (2ae^{-a} - a^2e^{-a})(0 - a) + a^2e^{-a}\] \[0 = -2a^2e^{-a} + a^3e^{-a} + a^2e^{-a}\] \[0 = a^3e^{-a} - a^2e^{-a}\] Factor out \(a^2e^{-a}\): \[a^2e^{-a}(a - 1) = 0\] Since \(a > 0\), we cannot have \(a^2e^{-a} = 0\), so we must have: \[a - 1 = 0\] Therefore: \[a = 1\] So, the value of \(a > 0\) such that the tangent line to the graph of \(f(x) = x^2 e^{-x}\) at \(x = a\) passes through the origin is \(a = 1\). |
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| doubao | First find the derivative of f(x) = x^2e^{-x}: f'(x) = (2x)e^{-x} - x^2e^{-x} = x(2-x)e^{-x}. The slope of the tangent line at x = a is f'(a) = a(2-a)e^{-a}. The equation of the tangent line at x = a is y - f(a) = f'(a)(x - a). Since the tangent line passes through the origin (0,0), we have 0 - f(a) = f'(a)(0 - a), that is -a^2e^{-a} = -a^2(2-a)e^{-a}, which simplifies to 1 = 2 - a, so a = 1. |
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| minimax | To find the value of \( a > 0 \) such that the tangent line to the graph of \( f(x) = x^2 e^{-x} \) at \( x = a \) passes through the origin, we need to follow these steps: 1. Find the derivative of \( f(x) \) to get the slope of the tangent line at any point \( x \). 2. Evaluate the derivative at \( x = a \) to get the slope of the tangent line at \( x = a \). 3. Use the point-slope form of the line equation to write the equation of the tangent line at \( x = a \). 4. Ensure that the tangent line passes through the origin (0,0), which will give us an equation to solve for \( a \). Let's go through these steps: 1. Find the derivative of \( f(x) \): \( f(x) = x^2 e^{-x} \) Using the product rule, where \( u = x^2 \) and \( v = e^{-x} \), we have: \( u' = 2x \) \( v' = -e^{-x} \) So the derivative \( f'(x) \) is: \( f'(x) = u'v + uv' \) \( f'(x) = (2x)e^{-x} + x^2(-e^{-x}) \) \( f'(x) = 2xe^{-x} - x^2e^{-x} \) 2. Evaluate the derivative at \( x = a \): \( f'(a) = 2ae^{-a} - a^2e^{-a} \) This is the slope of the tangent line at \( x = a \). 3. Write the equation of the tangent line at \( x = a \): The point-slope form of the line equation is: \( y - y_1 = m(x - x_1) \) Where \( m \) is the slope and \( (x_1, y_1) \) is a point on the line. Since we want the line to pass through the origin, \( (x_1, y_1) = (a, 0) \), and \( m = f'(a) \). So the equation becomes: \( y - 0 = f'(a)(x - a) \) \( y = f'(a)(x - a) \) Substituting \( f'(a) \) with the expression we found earlier: \( y = (2ae^{-a} - a^2e^{-a})(x - a) \) 4. Ensure the tangent line passes through the origin: Since the line passes through the origin, the y-intercept must be 0. This means that when \( x = 0 \), \( y = 0 \). We can use this to solve for \( a \): \( 0 = (2ae^{-a} - a^2e^{-a})(0 - a) \) \( 0 = -a(2ae^{-a} - a^2e^{-a}) \) Since \( e^{-a} \) is never zero, we can divide both sides by \( -e^{-a} \) to get: \( 0 = 2a^2 - a^3 \) Now we can factor out \( a^2 \): \( 0 = a^2(2 - a) \) This gives us two possible solutions for \( a \): \( a = 0 \) or \( a = 2 \) However, since we are looking for \( a > 0 \), we discard \( a = 0 \) and take \( a = 2 \) as the solution. Therefore, the value of \( a > 0 \) such that the tangent line to the graph of \( f(x) = x^2 e^{-x} \) at \( x = a \) passes through the origin is \( a = 2 \). |
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| kimi | To find the value of \( a > 0 \) such that the tangent line to the graph of \( f(x) = x^2 e^{-x} \) at \( x = a \) passes through the origin, we need to follow these steps: 1. **Find the derivative of the function \( f(x) \)** to determine the slope of the tangent line at any point \( x \). 2. **Evaluate the derivative at \( x = a \)** to find the slope of the tangent line at that specific point. 3. **Write the equation of the tangent line** using the point-slope form of a line. 4. **Set the y-intercept of the tangent line to zero** to ensure it passes through the origin. Let's start with step 1: The derivative of \( f(x) = x^2 e^{-x} \) can be found using the product rule, which states that \( (uv)' = u'v + uv' \), where \( u = x^2 \) and \( v = e^{-x} \). The derivative of \( u = x^2 \) is \( u' = 2x \). The derivative of \( v = e^{-x} \) is \( v' = -e^{-x} \). So, the derivative \( f'(x) \) is: \[ f'(x) = (x^2)' e^{-x} + x^2 (e^{-x})' = 2x e^{-x} - x^2 e^{-x} \] Simplifying, we get: \[ f'(x) = e^{-x}(2x - x^2) \] Now, for step 2, we evaluate the derivative at \( x = a \): \[ f'(a) = e^{-a}(2a - a^2) \] For step 3, the equation of the tangent line at \( x = a \) is: \[ y - f(a) = f'(a)(x - a) \] Since \( f(a) = a^2 e^{-a} \), the equation becomes: \[ y - a^2 e^{-a} = e^{-a}(2a - a^2)(x - a) \] For step 4, we set \( x = 0 \) and \( y = 0 \) to find \( a \): \[ 0 - a^2 e^{-a} = e^{-a}(2a - a^2)(0 - a) \] \[ -a^2 e^{-a} = -a e^{-a}(2a - a^2) \] Divide both sides by \( -e^{-a} \) (which is non-zero for \( a > 0 \)): \[ a^2 = a(2a - a^2) \] Simplify the equation: \[ a^2 = 2a^2 - a^3 \] Rearrange the terms: \[ a^3 - a^2 + a^2 = 0 \] \[ a^3 = 0 \] This simplifies to \( a = 0 \), which is not greater than 0. We made an error in our simplification. Let's go back to the step where we divide by \( -e^{-a} \) and simplify correctly: \[ a^2 = a(2a - a^2) \] \[ a^2 = 2a^2 - a^3 \] \[ 0 = a^3 - a^2 \] \[ 0 = a^2(a - 1) \] Since \( a > 0 \), the only solution is \( a = 1 \). So, the value of \( a \) such that the tangent line to the graph of \( f(x) = x^2 e^{-x} \) at \( x = a \) passes through the origin is \( a = 1 \). |
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| wenxinyiyan | To solve this problem, we need to find the equation of the tangent line to the graph of f(x) = x^2 * e^(-x) at x = a, and then determine the value of a such that this tangent line passes through the origin. 1. **Find the derivative of f(x)**: f'(x) = d/dx (x^2 * e^(-x)) Using the product rule: f'(x) = (2x * e^(-x)) + (x^2 * (-e^(-x))) f'(x) = e^(-x) * (2x - x^2) 2. **Find the slope of the tangent line at x = a**: m = f'(a) = e^(-a) * (2a - a^2) 3. **Find the equation of the tangent line**: Using the point-slope form of a line, y - y1 = m(x - x1), where (x1, y1) is the point on the graph of f(x) at x = a, so y1 = f(a) = a^2 * e^(-a), and m is the slope found in step 2. y - a^2 * e^(-a) = e^(-a) * (2a - a^2) * (x - a) 4. **Find the value of a such that the tangent line passes through the origin (0,0)**: Substitute x = 0, y = 0 into the equation of the tangent line: 0 - a^2 * e^(-a) = e^(-a) * (2a - a^2) * (0 - a) -a^2 * e^(-a) = -a * e^(-a) * (2a - a^2) Since e^(-a) ≠ 0 for any a, we can divide both sides by e^(-a): -a^2 = -a * (2a - a^2) a^2 = a * (2a - a^2) a^2 = 2a^2 - a^3 a^3 - a^2 = 0 a^2(a - 1) = 0 a = 0 or a = 1 But since the problem specifies that a > 0, we reject the solution a = 0. Therefore, the value of a such that the tangent line to the graph of f(x) = x^2 * e^(-x) at x = a passes through the origin is a = 1. |
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| zhipu | To find the value of \( a > 0 \) such that the tangent line to the graph of \( f(x) = x^2 e^{-x} \) at \( x = a \) passes through the origin, we need to follow these steps: 1. **Find the derivative of \( f(x) \) to get the slope of the tangent line at any point \( x \).** The function is \( f(x) = x^2 e^{-x} \). We use the product rule to find its derivative: \[ f'(x) = \frac{d}{dx}(x^2) \cdot e^{-x} + x^2 \cdot \frac{d}{dx}(e^{-x}) \] \[ f'(x) = 2x \cdot e^{-x} + x^2 \cdot (-e^{-x}) \] \[ f'(x) = 2x e^{-x} - x^2 e^{-x} \] \[ f'(x) = e^{-x} (2x - x^2) \] 2. **Evaluate the derivative at \( x = a \) to find the slope of the tangent line at \( x = a \).** \[ f'(a) = e^{-a} (2a - a^2) \] 3. **Find the equation of the tangent line at \( x = a \).** The general form of the equation of a line is \( y = mx + b \), where \( m \) is the slope and \( b \) is the y-intercept. Since the tangent line passes through the point \( (a, f(a)) \), we can write: \[ y - f(a) = f'(a) (x - a) \] Given \( f(a) = a^2 e^{-a} \) and \( f'(a) = e^{-a} (2a - a^2) \), the equation becomes: \[ y - a^2 e^{-a} = e^{-a} (2a - a^2) (x - a) \] 4. **Determine the condition for the tangent line to pass through the origin \( (0, 0) \).** Substitute \( (0, 0) \) into the equation of the tangent line: \[ 0 - a^2 e^{-a} = e^{-a} (2a - a^2) (0 - a) \] Simplify the equation: \[ -a^2 e^{-a} = -a e^{-a} (2a - a^2) \] \[ -a^2 = -a (2a - a^2) \] \[ a^2 = a (2a - a^2) \] Divide both sides by \( a \) (since \( a \neq 0 \)): \[ a = 2a - a^2 \] Rearrange the equation: \[ a^2 - a = 0 \] Factor the equation: \[ a(a - 1) = 0 \] This gives us two solutions: \( a = 0 \) or \( a = 1 \). Since the problem specifies \( a > 0 \), we discard \( a = 0 \). Therefore, the value of \( a \) is: \[ a = 1 \] So, the value of \( a > 0 \) such that the tangent line to the graph of \( f(x) = x^2 e^{-x} \) at \( x = a \) passes through the origin is \( \boxed{1} \). |
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| gpto1 | To find the value of \( a > 0 \) such that the tangent line to the graph of \( f(x) = x^2 e^{-x} \) at \( x = a \) passes through the origin \((0, 0)\), we proceed as follows. **Step 1: Find the equation of the tangent line at \( x = a \).** The equation of the tangent line to \( f(x) \) at \( x = a \) is given by: \[ y = f(a) + f'(a)(x - a) \] **Step 2: Set the tangent line to pass through the origin.** For the tangent line to pass through the origin \((0, 0)\): \[ 0 = f(a) + f'(a)(0 - a) \] Simplify: \[ 0 = f(a) - a f'(a) \] This leads to the equation: \[ a f'(a) = f(a) \] **Step 3: Compute \( f(a) \) and \( f'(a) \).** First, compute \( f(a) \): \[ f(a) = a^2 e^{-a} \] Next, compute \( f'(x) \) using the product rule: \[ f(x) = x^2 e^{-x} \] \[ f'(x) = \left(2x\right)e^{-x} + x^2(-e^{-x}) = e^{-x}\left(2x - x^2\right) \] So, \[ f'(a) = e^{-a}\left(2a - a^2\right) \] **Step 4: Substitute \( f(a) \) and \( f'(a) \) back into the equation.** Substitute \( f(a) \) and \( f'(a) \) into \( a f'(a) = f(a) \): \[ a \left[e^{-a}\left(2a - a^2\right)\right] = a^2 e^{-a} \] Simplify: \[ e^{-a} a \left(2a - a^2\right) = a^2 e^{-a} \] Subtract \( a^2 e^{-a} \) from both sides: \[ e^{-a} a \left(2a - a^2\right) - a^2 e^{-a} = 0 \] Factor out \( e^{-a} \): \[ e^{-a} \left[ a(2a - a^2) - a^2 \right] = 0 \] Simplify inside the brackets: \[ a(2a - a^2) - a^2 = 2a^2 - a^3 - a^2 = a^2 - a^3 \] So the equation becomes: \[ e^{-a} \left[ a^2 (1 - a) \right] = 0 \] **Step 5: Solve for \( a \).** Since \( e^{-a} > 0 \) for all real \( a \), we set: \[ a^2 (1 - a) = 0 \] This gives: \[ a^2 = 0 \quad \text{or} \quad 1 - a = 0 \] Solving these equations: 1. \( a^2 = 0 \) leads to \( a = 0 \). 2. \( 1 - a = 0 \) leads to \( a = 1 \). **Step 6: Consider the domain restriction.** Since \( a > 0 \), \( a = 0 \) is not acceptable. Therefore, the only valid solution is \( a = 1 \). **Answer:** \[ a = 1 \] |
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