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Problem Details

1. Function definitions:

   • $g(x) = \frac{x - 9}{11}$
   • $h = \{(-6, 5), (-3, -6), (5, 8), (8, 7)\}$

2. Tasks:

   • Find the inverse of $g$ and evaluate $g^{-1}(x)$.
   • Calculate $\left(g^{-1} \circ g\right)(-2)$.
   • Find $h^{-1}(5)$.

Step-by-Step Solutions

1. Find $g^{-1}(x)$

To find the inverse $g^{-1}(x)$, we’ll solve for $x$ in terms of $y$ by setting $y = g(x)$.

1. Set $y = \frac{x - 9}{11}$.
2. Solve for $x$ in terms of $y$: $y = \frac{x - 9}{11}$ Multiply both sides by 11: $11y = x - 9$ Add 9 to both sides: $x = 11y + 9$
3. Substitute $y$ back with $x$ to get $g^{-1}(x)$: $g^{-1}(x) = 11x + 9$

So, $g^{-1}(x) = 11x + 9$.

2. Find $\left(g^{-1} \circ g\right)(-2)$

This expression means we first apply $g$ to $-2$ and then apply $g^{-1}$ to the result.

1. Calculate $g(-2)$: $g(-2) = \frac{-2 - 9}{11} = \frac{-11}{11} = -1$
2. Now apply $g^{-1}$ to $-1$: $g^{-1}(-1) = 11(-1) + 9 = -11 + 9 = -2$

Thus, $\left(g^{-1} \circ g\right)(-2) = -2$.

3. Find $h^{-1}(5)$

To find $h^{-1}(5)$, look for the $x$-value in the set of points for $h$ where $h(x) = 5$.

From the set $h = \{(-6, 5), (-3, -6), (5, 8), (8, 7)\}$, we see that:

• When $x = -6$, $h(x) = 5$.

Thus, $h^{-1}(5) = -6$.

Answers

1. $g^{-1}(x) = 11x + 9$
2. $\left(g^{-1} \circ g\right)(-2) = -2$
3. $h^{-1}(5) = -6$

Would you like a more detailed explanation for any of these steps?

Related Questions

1. How do you determine if a function is one-to-one and has an inverse?
2. Can you explain the importance of domain and range when finding inverses?
3. How would you find $h^{-1}(x)$ if more values of $h$ were given?
4. Why does $\left(g^{-1} \circ g\right)(x) = x$ for all $x$ in the domain of $g$?
5. What are the steps to verify if an inverse function is correct?

Tip

Always verify your inverse by composing it with the original function. If $g(g^{-1}(x)) = x$ and $g^{-1}(g(x)) = x$, the inverse is correct.