To find the inverse of the given relation $y = \frac{\sqrt{x} + 1}{3}$, let's go through the steps and then determine the domain and range.

Step 1: Rewrite the equation

We start with the equation: $y = \frac{\sqrt{x} + 1}{3}$

Step 2: Solve for $x$ in terms of $y$

Multiply both sides by 3 to eliminate the fraction: $3y = \sqrt{x} + 1$

Subtract 1 from both sides: $3y - 1 = \sqrt{x}$

Now square both sides to remove the square root: $(3y - 1)^2 = x$

So, the inverse of the function is: $x = (3y - 1)^2$ Or, equivalently, the inverse relation is: $y^{-1}(x) = (3x - 1)^2$

Step 3: Determine the domain and range

• For the original function $y = \frac{\sqrt{x} + 1}{3}$, the square root term $\sqrt{x}$ requires $x \geq 0$. Therefore, the domain of the original function is $[0, \infty)$.

  The square root ensures that the output is non-negative, and dividing by 3 does not affect the positivity, so the range of the original function is $[\frac{1}{3}, \infty)$.

• For the inverse function $y^{-1}(x) = (3x - 1)^2$, we must ensure that the expression inside the square is non-negative. The domain of the inverse function is $[\frac{1}{3}, \infty)$, and since squaring any real number gives a non-negative result, the range of the inverse is $[0, \infty)$.

Summary:

• The inverse relation is: $y^{-1}(x) = (3x - 1)^2$
• Domain of the original function: $[0, \infty)$
• Range of the original function: $\left[ \frac{1}{3}, \infty \right)$
• Domain of the inverse function: $\left[ \frac{1}{3}, \infty \right)$
• Range of the inverse function: $[0, \infty)$

Do you have any questions or want more details?

Related Questions:

1. How do you find the inverse of a logarithmic function?
2. What are the conditions for a function to have an inverse?
3. Can all square root functions have inverses? Why or why not?
4. How does the domain of a function affect its inverse?
5. How can you determine if a relation is a function?

Tip: Always ensure that the function has a one-to-one correspondence before attempting to find its inverse.