Question

Given

f (x) = √√x + 2 − 1.
√√x+2 -1, what is the relationship between
Drag and drop an inequality into each box to correctly complete each statement.
x > -2 x ≥ −1 x ≥ 0
The domain of
The range of
f (x) is
f (x) is
31
x ≥0 y≥-2 y≥-1 y 20
so the range of
f (x) and f-¹ (x)₂
?
so the domain of
ƒ-¹ (x) is
f-
ƒ-¹ (x) is
4
1 2
3 4
5
Finish ▶
Sep 6
12:31 O

Answer 1

Explanation:

Let's analyze the given function \( f(x) = \sqrt{\sqrt{x + 2} - 1} \) and determine its domain and range, as well as the domain and range of its inverse function \( f^{-1}(x) \).

1. Domain of \( f(x) \):

- The expression inside the square root, \( \sqrt{x + 2} - 1 \), must be greater than or equal to 0 to ensure real values under the square root.

- So, \( x + 2 - 1 \geq 0 \).

- Solve for \( x \):

\( x + 1 \geq 0 \)

\( x \geq -1 \)

Therefore, the domain of \( f(x) \) is \( x \geq -1 \).

2. Range of \( f(x) \):

- To find the range, we need to consider the possible output values of \( f(x) \).

- Since the square root of a non-negative number is always non-negative, \( \sqrt{\text{anything}} \) is always non-negative.

- Subtracting 1 from a non-negative number keeps it non-negative.

So, the range of \( f(x) \) is \( y \geq 0 \).

3. Domain of \( f^{-1}(x) \):

- To find the domain of the inverse function, we need to consider the possible input values.

- The range of \( f(x) \) is \( y \geq 0 \), which means the input values for \( f^{-1}(x) \) should be \( x \geq 0 \).

Therefore, the domain of \( f^{-1}(x) \) is \( x \geq 0 \).

4. Range of \( f^{-1}(x) \):

- The range of the inverse function \( f^{-1}(x) \) corresponds to the domain of the original function \( f(x) \).

- So, the range of \( f^{-1}(x) \) is \( y \geq -1 \).

Now, let's fill in the statements:

- The domain of \( f(x) \) is \( x \geq -1 \).

- The range of \( f(x) \) is \( y \geq 0 \).

- The range of \( f^{-1}(x) \) is \( y \geq -1 \).

- The domain of \( f^{-1}(x) \) is \( x \geq 0 \).