Explain the range of f (x)=sqrt(x-1) by equating it to y

To find the range of the function \( f(x) = \sqrt{x - 1} \), we can start by setting it equal to \( y \):

Setting Up the Equation

We rewrite the function as:

y = √(x - 1)

Isolating x

Next, we can square both sides to eliminate the square root:

y² = x - 1

Now, we can solve for \( x \):

x = y² + 1

Determining Valid Values for y

Since \( f(x) \) involves a square root, the expression inside the square root must be non-negative:

x - 1 ≥ 0

This means:

x ≥ 1

Finding the Range

Now, substituting back, we see that as \( x \) starts from 1 and increases, \( y \) will take on values starting from:

y = √(1 - 1) = 0

As \( x \) increases, \( y \) can take any non-negative value. Thus, the range of \( f(x) \) is:

• y ≥ 0

Final Thoughts

In summary, the range of the function \( f(x) = \sqrt{x - 1} \) is all non-negative real numbers, or in interval notation, it is:

[0, ∞)