Question

question 13
let $f(x) = 4 + \sqrt{5x - 8}$. find $f^{-1}(x)$.
$f^{-1}(x) = \square$
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Explanation:

Step1: Swap \(x\) and \(y\)

Let \(y = f(x)=4+\sqrt{5x - 8}\). Swap \(x\) and \(y\) to get \(x = 4+\sqrt{5y - 8}\).

Step2: Isolate the square root

Subtract 4 from both sides: \(x - 4=\sqrt{5y - 8}\).

Step3: Square both sides

Square both sides to eliminate the square root: \((x - 4)^2=5y - 8\).

Step4: Solve for \(y\)

First, expand \((x - 4)^2\) to get \(x^2-8x + 16=5y - 8\). Then add 8 to both sides: \(x^2-8x + 24=5y\). Finally, divide both sides by 5: \(y=\frac{(x - 4)^2 + 8}{5}=\frac{x^2-8x + 16 + 8}{5}=\frac{x^2-8x + 24}{5}\). Also, we need to consider the domain of the inverse function. The original function \(f(x)\) has a range \(y\geq4\) (since square root is non - negative and we add 4), so the domain of \(f^{-1}(x)\) is \(x\geq4\).

Answer:

\(f^{-1}(x)=\frac{(x - 4)^2 + 8}{5}=\frac{x^{2}-8x + 24}{5}\) (for \(x\geq4\))