Question

chapter 1 quiz (1.1,1.2, 1.4, 1.5, 1.7)
14 points possible answered: 12/14
question 13
let $f(x) = 9 + \sqrt{6x - 7}$. find $f^{-1}(x)$.
$f^{-1}(x) = $

Explanation:

Step1: Swap x and y

Let \( y = 9 + \sqrt{6x - 7} \). Swap \( x \) and \( y \) to get \( x = 9 + \sqrt{6y - 7} \).

Step2: Isolate the square root

Subtract 9 from both sides: \( x - 9 = \sqrt{6y - 7} \).

Step3: Square both sides

Square both sides to eliminate the square root: \( (x - 9)^2 = 6y - 7 \).

Step4: Solve for y

Add 7 to both sides: \( (x - 9)^2 + 7 = 6y \). Then divide both sides by 6: \( y = \frac{(x - 9)^2 + 7}{6} \). Simplify the numerator: \( (x - 9)^2 = x^2 - 18x + 81 \), so \( y = \frac{x^2 - 18x + 81 + 7}{6} = \frac{x^2 - 18x + 88}{6} \). We also need to consider the domain of the inverse function. The original function \( f(x) \) has a range \( y \geq 9 \) (since the square root is non - negative and we add 9), so the domain of \( f^{-1}(x) \) is \( x \geq 9 \).

Answer:

\( f^{-1}(x)=\frac{(x - 9)^2 + 7}{6}=\frac{x^{2}-18x + 88}{6}\) (for \( x\geq9 \))