Question
upstudy study bank question image url

Find the inverse of the given square root equation. \( \begin{array}{l}\text { 4. } f(x)=2 \sqrt{x-1}+5\end{array} \)

Ask by Romero Harris. in the United States
Jan 23,2025

Real Tutor Solution

Tutor-Verified Answer

Answer

The inverse function is \( f^{-1}(x) = \left( \frac{x - 5}{2} \right)^2 + 1 \).

Solution

To find the inverse of the function \( f(x) = 2 \sqrt{x - 1} + 5 \), follow these steps: 1. **Start with the original equation:** \[ y = 2 \sqrt{x - 1} + 5 \] 2. **Isolate the square root term:** \[ y - 5 = 2 \sqrt{x - 1} \] 3. **Divide both sides by 2:** \[ \frac{y - 5}{2} = \sqrt{x - 1} \] 4. **Square both sides to eliminate the square root:** \[ \left( \frac{y - 5}{2} \right)^2 = x - 1 \] 5. **Solve for \( x \):** \[ x = \left( \frac{y - 5}{2} \right)^2 + 1 \] 6. **Express the inverse function by swapping \( x \) and \( y \):** \[ f^{-1}(x) = \left( \frac{x - 5}{2} \right)^2 + 1 \] So, the inverse function is: \[ f^{-1}(x) = \left( \frac{x - 5}{2} \right)^2 + 1 \] **Final Answer:** \[ f^{-1}(x) = \left(\frac{x - 5}{2}\right)^{\,2} + 1 \]

Reviewed and approved by the UpStudy tutoring team

error msg
Explain
Simplify this solution

Extra Insights

To find the inverse of the function \( f(x) = 2 \sqrt{x - 1} + 5 \), we start by replacing \( f(x) \) with \( y \): \( y = 2 \sqrt{x - 1} + 5 \). Next, isolate the square root by subtracting 5 from both sides: \( y - 5 = 2 \sqrt{x - 1} \). Now, divide by 2: \( \frac{y - 5}{2} = \sqrt{x - 1} \). To eliminate the square root, square both sides: \( \left(\frac{y - 5}{2}\right)^2 = x - 1 \). Finally, solve for \( x \): \( x = \left(\frac{y - 5}{2}\right)^2 + 1 \). Thus, the inverse function is: \( f^{-1}(y) = \left(\frac{y - 5}{2}\right)^2 + 1 \). Replacing \( y \) back with \( x \) gives: \( f^{-1}(x) = \left(\frac{x - 5}{2}\right)^2 + 1 \). Isn’t it cool how we can reverse functions like this? It’s like flipping a pancake – you get to see the other side! When working with square roots, remember to check for any extraneous solutions. Often, squaring both sides can introduce values that don’t satisfy the original equation, so it’s always good to plug your inverse back into the original function to ensure they match up correctly!

Try Premium now!
Try Premium and ask Thoth AI unlimited math questions now!
Maybe later Go Premium
Study can be a real struggle
Why not UpStudy it?
Select your plan below
Premium

You can enjoy

Start now
  • Step-by-step explanations
  • 24/7 expert live tutors
  • Unlimited number of questions
  • No interruptions
  • Full access to Answer and Solution
  • Full Access to PDF Chat, UpStudy Chat, Browsing Chat
Basic

Totally free but limited

  • Limited Solution
Welcome to UpStudy!
Please sign in to continue the Thoth AI Chat journey
Continue with Email
Or continue with
By clicking “Sign in”, you agree to our Terms of Use & Privacy Policy