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Let \( f(x)=4+\sqrt{6 x-7} \). Find \( f^{-1}(x) \) \( f^{-1}(x)=\square \), for \( x \geq 4 \)

Ask by Pope Fuentes. in the United States
Mar 15,2025

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\( f^{-1}(x) = \frac{(x-4)^2 + 7}{6} \), for \( x \geq 4 \).

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The Deep Dive

To find the inverse of the function \( f(x) = 4 + \sqrt{6x - 7} \), we start by setting \( y = f(x) \). This gives us the equation: \[ y = 4 + \sqrt{6x - 7} \] Next, we will solve for \( x \) in terms of \( y \). First, subtract 4 from both sides: \[ y - 4 = \sqrt{6x - 7} \] Now, square both sides to eliminate the square root: \[ (y - 4)^2 = 6x - 7 \] Next, rearrange to isolate \( x \): \[ 6x = (y - 4)^2 + 7 \] \[ x = \frac{(y - 4)^2 + 7}{6} \] Now, replacing \( y \) with \( x \) in the inverse function, we have: \[ f^{-1}(x) = \frac{(x - 4)^2 + 7}{6} \] Thus, the final expression for the inverse function is: \[ f^{-1}(x) = \frac{(x - 4)^2 + 7}{6}, \text{ for } x \geq 4 \]

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