Question

homework assignment 1.7: inverse functions
score: 9.5/15 answered: 10/15
question 11
let $f(x) = 5 + \sqrt{6x - 3}$. find $f^{-1}(x)$.
$f^{-1}(x) = $
question help: video written example

Explanation:

Step1: Swap x and y

Let \( y = 5+\sqrt{6x - 3} \). Swap \( x \) and \( y \), we get \( x = 5+\sqrt{6y - 3} \).

Step2: Isolate the square root term

Subtract 5 from both sides: \( x - 5=\sqrt{6y - 3} \).

Step3: Square both sides

Square both sides to eliminate the square root: \( (x - 5)^2=6y - 3 \).

Step4: Solve for y

First, expand \( (x - 5)^2 \): \( x^{2}-10x + 25=6y - 3 \). Then add 3 to both sides: \( x^{2}-10x + 28=6y \). Finally, divide both sides by 6: \( y=\frac{x^{2}-10x + 28}{6} \). We can also simplify it as \( y=\frac{x^{2}}{6}-\frac{5x}{3}+\frac{14}{3} \). Also, we need to consider the domain of the inverse function. The original function \( f(x) \) has a range \( y\geq5 \) (since the square root is non - negative, \( \sqrt{6x - 3}\geq0 \), so \( y = 5+\sqrt{6x - 3}\geq5 \)), so the domain of \( f^{-1}(x) \) is \( x\geq5 \).

Answer:

\( f^{-1}(x)=\frac{x^{2}-10x + 28}{6} \) (or \( \frac{x^{2}}{6}-\frac{5x}{3}+\frac{14}{3} \)) with \( x\geq5 \)