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Let's solve the problem step by step. We are given the function \( f(x) = \frac{x^2 - 3}{2} \).

### Part (a): Find \( f(3) \)
To find \( f(3) \), substitute \( x = 3 \) into the given function.

\[
f(3) = \frac{3^2 - 3}{2} = \frac{9 - 3}{2} = \frac{6}{2} = 3
\]

So, \( f(3) = 3 \).

### Part (b): Find \( f^{-1}(x) \)
To find the inverse function \( f^{-1}(x) \), start by setting \( y = f(x) \) and solve for \( x \) in terms of \( y \).

\[
y = \frac{x^2 - 3}{2}
\]

Multiply both sides by 2 to eliminate the fraction:

\[
2y = x^2 - 3
\]

Next, add 3 to both sides:

\[
x^2 = 2y + 3
\]

Finally, solve for \( x \):

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

Since the function \( f(x) \) is quadratic, its inverse will have two branches, corresponding to the positive and negative square roots. For simplicity, let's take the principal branch (positive root):

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

### Part (c): Find \( f^{-1}(11) \)
Now, substitute \( x = 11 \) into the inverse function:

\[
f^{-1}(11) = \sqrt{2(11) + 3} = \sqrt{22 + 3} = \sqrt{25} = 5
\]

So, \( f^{-1}(11) = 5 \).

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If you have any questions or need further details, feel free to ask!

Here are 5 related questions you can explore:
1. What would the inverse function look like if we considered the negative square root?
2. How would you determine the domain and range of the inverse function \( f^{-1}(x) \)?
3. What are the domain and range of the original function \( f(x) \)?
4. How would you verify that \( f(f^{-1}(x)) = x \) for the principal branch?
5. What is the significance of choosing the principal branch in the context of inverse functions?

**Tip:** When dealing with inverse functions, always remember to check the domain and range of the original function to ensure that the inverse is valid over the desired interval.

Given that f(x) = (x^2 - 3) / 2, a) Find f(3), b) Find f^(-1)(x), c) Find f^(-1)(11)

Learn how to find f(3), the inverse function f^(-1)(x), and f^(-1)(11) for f(x) = (x^2 - 3) / 2. Explore quadratic equations and inverse functions in this detailed, step-by-step solution suitable for high school students.

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