Question

the function \\(f(x) = x + 7\\) is one-to-one.

a. find an equation for \\(f^{-1}(x)\\), the inverse function.
b. verify that your equation is correct by showing that \\(f(f^{-1}(x)) = x\\) and \\(f^{-1}(f(x)) = x\\).

a. \\(f^{-1}(x) = \quad\\), for \\(x \le \quad\\)
b. \\(f^{-1}(x) = x - 7\\), for all \\(x\\)
c. \\(f^{-1}(x) = \quad\\), for \\(x \ge \quad\\)
d. \\(f^{-1}(x) = \quad\\), for \\(x \
e \quad\\)

b. verify that the equation is correct.
\\(f(f^{-1}(x)) = f(x - 7)\\) and \\(f^{-1}(f(x)) = f^{-1}(x + 7)\\) substitute.
\\(= x\\) and \\(= x\\) simplify.

the equation is

Explanation:

Find the inverse function

Using the Inverse Functions knowledge point:
We set \(y = f(x)\):

$$y = x + 7$$

Swap \(x\) and \(y\) to find the inverse:

$$x = y + 7$$

Solve for \(y\):

$$y = x - 7$$

Thus, \(f^{-1}(x) = x - 7\). Since the domain of \(f(x)\) is all real numbers, the domain of \(f^{-1}(x)\) is also all real numbers. This corresponds to option B: \(f^{-1}(x) = x - 7\), for all \(x\).

Verify the composition

Using the Function Composition knowledge point:
We compute both compositions to verify:

$$f(f^{-1}(x)) = f(x - 7) = (x - 7) + 7 = x$$

$$f^{-1}(f(x)) = f^{-1}(x + 7) = (x + 7) - 7 = x$$

Since both compositions simplify to \(x\), the equation is verified.

Answer:

Question 1

• (A) \(f^{-1}(x) = \text{blank}\), for \(x \le \text{blank}\)
• (B) \(f^{-1}(x) = x - 7\), for all \(x\) (Correct answer)
• (C) \(f^{-1}(x) = \text{blank}\), for \(x \ge \text{blank}\)
• (D) \(f^{-1}(x) = \text{blank}\), for \(x

e \text{blank}\)

Question 2

The equation is <blank>verified</blank>.