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7. Find the inverse of the function \( f(x)=\frac{1}{5} x-2 \) by switching the roles of \( x \) and \( y \) and solving for \( y \). Then find the inverse of the function \( f \) by using inverse operations in the reverse order. Which method do you prefer? Explain.

Ask by Hobbs Ruiz. in the United States
Mar 17,2025

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Answer

要找到函数 \( f(x) = \frac{1}{5} x - 2 \) 的反函数,可以使用两种方法: 1. **交换角色法**: - 将 \( y = \frac{1}{5} x - 2 \) 中的 \( x \) 和 \( y \) 交换,得到 \( x = \frac{1}{5} y - 2 \)。 - 解这个方程,得到 \( y = 5x + 10 \)。 - 所以反函数是 \( f^{-1}(x) = 5x + 10 \)。 2. **逆操作法**: - 从 \( y = \frac{1}{5} x - 2 \) 开始,先加 2,再乘以 5,得到 \( x = 5y + 10 \)。 - 交换 \( x \) 和 \( y \),得到 \( y = 5x + 10 \)。 - 反函数同样是 \( f^{-1}(x) = 5x + 10 \)。 **个人偏好**:我更喜欢第一种方法,因为它步骤清晰,容易理解。逆操作法虽然有效,但需要更多的步骤和注意,容易出错。

Solution

要找到函数 \( f(x) = \frac{1}{5} x - 2 \) 的反函数,我们可以通过交换 \( x \) 和 \( y \) 的角色并解出 \( y \) 来进行。 ### 方法一:交换角色法 1. 首先,将 \( f(x) \) 表示为 \( y \): \[ y = \frac{1}{5} x - 2 \] 2. 交换 \( x \) 和 \( y \): \[ x = \frac{1}{5} y - 2 \] 3. 解这个方程以找到 \( y \): - 首先,加上 2: \[ x + 2 = \frac{1}{5} y \] - 然后,乘以 5: \[ 5(x + 2) = y \] - 最终得到: \[ y = 5x + 10 \] 因此,反函数为: \[ f^{-1}(x) = 5x + 10 \] ### 方法二:逆操作法 我们也可以通过逆操作的方式来找到反函数。我们从 \( f(x) \) 开始,逐步应用逆操作。 1. 从 \( y = \frac{1}{5} x - 2 \) 开始。 2. 首先,加上 2: \[ y + 2 = \frac{1}{5} x \] 3. 然后,乘以 5: \[ 5(y + 2) = x \] 4. 最终得到: \[ x = 5y + 10 \] 5. 交换 \( x \) 和 \( y \): \[ y = 5x + 10 \] 因此,反函数同样为: \[ f^{-1}(x) = 5x + 10 \] ### 结论 两种方法都得到了相同的反函数 \( f^{-1}(x) = 5x + 10 \)。我个人更喜欢第一种方法(交换角色法),因为它在逻辑上更直接,步骤清晰,容易理解。逆操作法虽然也有效,但可能需要更多的步骤和注意力来确保每一步都正确。

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To find the inverse of the function \( f(x)=\frac{1}{5} x-2 \), start by switching \( x \) and \( y \): 1. Rewrite the function as \( y = \frac{1}{5} x - 2 \). 2. Switch the variables: \( x = \frac{1}{5} y - 2 \). 3. Solve for \( y \): \[ x + 2 = \frac{1}{5} y \] \[ y = 5(x + 2) = 5x + 10. \] Thus, \( f^{-1}(x) = 5x + 10 \). Using inverse operations, start with \( y = f(x) \): 1. Add 2 to both sides: \( y + 2 = \frac{1}{5} x \). 2. Multiply by 5: \( 5(y + 2) = x \). 3. Rearranging gives \( 5y + 10 = x \), leading to \( y = \frac{x - 10}{5} \). This confirms our earlier result. The first method feels more direct as it provides a clear algebraic pathway without unnecessary manipulation of the original equation. It can be like switching dance partners—much easier to find the right steps!

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