Question

1. (20 points) determine if the function is 1-1. if the function is 1-1, find its inverse.

(a) (f(x) = 5x - 9)
(f^{-1}(x) = )

(b) (f(x) = |x - 2|)
(f^{-1}(x) = )

(c) (f(x) = \frac{2x + 3}{-5x + 8})
(f^{-1}(x) = )

(d) (f(x) = \sqrt5{x} - 2)
(f^{-1}(x) = )

Explanation:

Step1: Analyze part (a)

\(f(x) = 5x - 9\) is a linear function, which is always 1-1.

$$y = 5x - 9 \implies x = \frac{y + 9}{5} \implies f^{-1}(x) = \frac{x + 9}{5}$$

Step2: Analyze part (b)

\(f(x) = |x - 2|\) is an absolute value function.

$$f(1) = |1 - 2| = 1, \quad f(3) = |3 - 2| = 1$$

Since \(f(1) = f(3)\) for \(1
eq 3\), it is not 1-1.

Step3: Analyze part (c)

\(f(x) = \frac{2x + 3}{-5x + 8}\) is a rational function, which is 1-1 on its domain.

$$y = \frac{2x + 3}{-5x + 8} \implies y(-5x + 8) = 2x + 3 \implies x(5y + 2) = 8y - 3 \implies f^{-1}(x) = \frac{8x - 3}{5x + 2}$$

Step4: Analyze part (d)

\(f(x) = \sqrt[5]{x} - 2\) is a strictly increasing odd-root function, which is 1-1.

$$y = \sqrt[5]{x} - 2 \implies y + 2 = \sqrt[5]{x} \implies x = (y + 2)^5 \implies f^{-1}(x) = (x + 2)^5$$

Answer:

(a) The function is 1-1; \(f^{-1}(x) = \frac{x + 9}{5}\)
(b) The function is not 1-1.
(c) The function is 1-1; \(f^{-1}(x) = \frac{8x - 3}{5x + 2}\)
(d) The function is 1-1; \(f^{-1}(x) = (x + 2)^5\)