414,901 views
32 votes
32 votes
The one-to-one functions 9 and h are defined as follows.g={(0, 5), (2, 4), (4, 6), (5, 9), (9, 0)}h(x)X +811

The one-to-one functions 9 and h are defined as follows.g={(0, 5), (2, 4), (4, 6), (5, 9), (9, 0)}h-example-1
User Jatin Gupta
by
2.9k points

1 Answer

28 votes
28 votes

Step 1: Write out the functions

g(x) = { (0.5), (2, 4), (4,6), (5,9), (9,0) }


h(x)\text{ = }\frac{x\text{ + 8}}{11}

Step 2:

For the function g(x),

The inputs variables are: 0 , 2, 4, 5, 9

The outputs variables are: 5, 4, 6, 9, 0

The inverse of an output is its input value.

Therefore,


g^(-1)(9)\text{ = 5}

Step 3: find the inverse of h(x)

To find the inverse of h(x), let y = h(x)


\begin{gathered} h(x)\text{ = }\frac{x\text{ + 8}}{11} \\ y\text{ = }\frac{x\text{ + 8}}{11} \\ \text{Cross multiply} \\ 11y\text{ = x + 8} \\ \text{Make x subject of formula} \\ 11y\text{ - 8 = x} \\ \text{Therefore, h}^(-1)(x)\text{ = 11x - 8} \\ h^(-1)(x)\text{ = 11x - 8} \end{gathered}

Step 4:


Find(h.h^(-1))(1)
\begin{gathered} h(x)\text{ = }\frac{x\text{ + 8}}{11} \\ h^(-1)(x)\text{ = 11x - 8} \\ \text{Next, substitute h(x) inverse into h(x).} \\ \text{Therefore} \\ (h.h^(-1))\text{ = }\frac{11x\text{ - 8 + 8}}{11} \\ h.h^(-1)(x)\text{ = x} \\ h.h^(-1)(1)\text{ = 1} \end{gathered}

Step 5: Final answer


\begin{gathered} g^(-1)(9)\text{ = 5} \\ h^(-1)(x)\text{ = 11x - 8} \\ h\lbrack h^(-1)(x)\rbrack\text{ = 1} \end{gathered}

User Dela
by
2.7k points