To find the inverse of the function $f(x) = \sqrt{x^2 + 8x}$, where $x > 0$, follow these steps:

Step 1: Write the function in terms of $y$

$y = \sqrt{x^2 + 8x}$

Step 2: Square both sides to eliminate the square root

$y^2 = x^2 + 8x$

Step 3: Rearrange the equation into a standard quadratic form

Rewriting the equation: $x^2 + 8x - y^2 = 0$ This is a quadratic equation in $x$.

Step 4: Solve the quadratic equation for $x$

We use the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ For the equation $x^2 + 8x - y^2 = 0$, $a = 1$, $b = 8$, and $c = -y^2$. Plugging these into the quadratic formula: $x = \frac{-8 \pm \sqrt{8^2 - 4(1)(-y^2)}}{2(1)}$ $x = \frac{-8 \pm \sqrt{64 + 4y^2}}{2}$ $x = \frac{-8 \pm \sqrt{4(y^2 + 16)}}{2}$ $x = \frac{-8 \pm 2\sqrt{y^2 + 16}}{2}$ $x = -4 \pm \sqrt{y^2 + 16}$

Step 5: Choose the correct sign

Since $x > 0$, we take the positive root: $x = -4 + \sqrt{y^2 + 16}$

Step 6: Write the inverse function

Therefore, the inverse function is: $f^{-1}(y) = -4 + \sqrt{y^2 + 16}$

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Would you like any further details or clarifications on this?

Here are 5 related questions for deeper understanding:

1. How do you verify that this function is indeed the inverse?
2. What happens if $x$ were allowed to take negative values?
3. How would the process change if the function were more complex?
4. Can you graph both the function and its inverse for better understanding?
5. How does the domain and range of the inverse relate to the original function?

Tip: To check if a function and its inverse are correct, compose them and see if the result simplifies to $x$.