Questions: Let f(x)=sqrt(2x-1)-3. Its inverse will have a domain of [-3, infinity). Which of the following represents f^(-1)(x) ? Select the correct answer below: f^(-1)(x)=(2x-1)^2+3 f^(-1)(x)=1/3(2x-1)^2 f^(-1)(x)=2(x+3)^2-1 f^(-1)(x)=1/2[(x+3)^2+1]

Let f(x)=sqrt(2x-1)-3. Its inverse will have a domain of [-3, infinity). Which of the following represents f^(-1)(x) ?

Select the correct answer below:
f^(-1)(x)=(2x-1)^2+3
f^(-1)(x)=1/3(2x-1)^2
f^(-1)(x)=2(x+3)^2-1
f^(-1)(x)=1/2[(x+3)^2+1]

Transcript text: Let $f(x)=\sqrt{2 x-1}-3$. Its inverse will have a domain of $[-3, \infty)$. Which of the following represents $f^{-1}(x)$ ? Select the correct answer below: $f^{-1}(x)=(2 x-1)^{2}+3$ $f^{-1}(x)=\frac{1}{3}(2 x-1)^{2}$ $f^{-1}(x)=2(x+3)^{2}-1$ $f^{-1}(x)=\frac{1}{2}\left[(x+3)^{2}+1\right]$

Solution

Solution Steps

Step 1: Define the Function

The original function is given by

\[ f(x) = \sqrt{2x - 1} - 3 \]

Step 2: Swap Variables

To find the inverse function, we swap \(x\) and \(y\):

\[ x = \sqrt{2y - 1} - 3 \]

Step 3: Isolate the Square Root

Next, we isolate the square root by adding 3 to both sides:

\[ x + 3 = \sqrt{2y - 1} \]

Step 4: Square Both Sides

Now, we square both sides to eliminate the square root:

\[ (x + 3)^2 = 2y - 1 \]

Step 5: Solve for \(y\)

We then solve for \(y\) by adding 1 to both sides and dividing by 2:

\[ 2y = (x + 3)^2 + 1 \] \[ y = \frac{1}{2} \left[(x + 3)^2 + 1\right] \]

Final Answer

Thus, the inverse function is

\[ \boxed{f^{-1}(x) = \frac{1}{2} \left[(x + 3)^2 + 1\right]} \]

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