Question

01-22 square root test
find the inverse of the following function.
$f(x) = 2(x - 3)^2 + 4$, for $x \leq 3$
$f^{-1}(x) = \boldsymbol{\text{dropdown}} \boldsymbol{\text{dropdown}} \sqrt{\boldsymbol{\text{dropdown}} \boldsymbol{\text{dropdown}}} \boldsymbol{\text{dropdown}}$

Explanation:

Step1: Replace \( f(x) \) with \( y \)

Let \( y = 2(x - 3)^2 + 4 \).

Step2: Swap \( x \) and \( y \)

We get \( x = 2(y - 3)^2 + 4 \).

Step3: Solve for \( y \)

First, subtract 4 from both sides:
\( x - 4 = 2(y - 3)^2 \)
Then, divide both sides by 2:
\( \frac{x - 4}{2} = (y - 3)^2 \)
Since the original function has \( x \leq 3 \), the inverse function will have a domain related to the range of the original function. The original function \( f(x)=2(x - 3)^2+4 \) with \( x\leq3 \) is a parabola opening upwards with vertex at \( (3,4) \) and decreasing for \( x\leq3 \), so the inverse function should be the left - hand side of the parabola (since we need to ensure the inverse is a function). Taking the square root of both sides, we consider the negative square root (because when we solved for \( y \), we want the part that corresponds to the original function's domain \( x\leq3 \)):
\( y - 3=-\sqrt{\frac{x - 4}{2}} \)
Then, add 3 to both sides:
\( y = 3-\sqrt{\frac{x - 4}{2}} \)
So \( f^{-1}(x)=3-\sqrt{\frac{x - 4}{2}} \)

Answer:

\( f^{-1}(x)=3-\sqrt{\frac{x - 4}{2}} \)