Question

the one-to-one functions \\(g\\) and \\(h\\) are defined as follows.

\\g = \\{(-4, 7), (-3, 6), (6, 0), (8, -6)\\}\\
\\h(x) = 2x - 9\\

find the following.

\\g^{-1}(6) = \\
\\h^{-1}(x) = \\
\\(h \circ h^{-1})(3) = \\

Explanation:

Find the inverse value of g

Using the Function Evaluation knowledge point

$$ \text{Since } g = \{(-4, 7), (-3, 6), (6, 0), (8, -6)\}, \text{ we have } g(-3) = 6. $$

$$ \text{Therefore, } g^{-1}(6) = -3. $$

Find the inverse function of h

To find \(h^{-1}(x)\), we set \(y = h(x)\) and solve for \(x\):

$$ y = 2x - 9 $$

$$ y + 9 = 2x \implies x = \frac{y + 9}{2} $$

$$ h^{-1}(x) = \frac{x + 9}{2} $$

Evaluate the composition of h and its inverse

Using the Function Composition knowledge point

$$ \text{For any } x \text{ in the domain of } h^{-1}, \text{ the identity holds: } (h \circ h^{-1})(x) = x. $$

$$ (h \circ h^{-1})(3) = 3 $$

Answer:

Question 1

\(g^{-1}(6) = -3\)

Question 2

\(h^{-1}(x) = \frac{x+9}{2}\)

Question 3

\((h \circ h^{-1})(3) = 3\)