Question: The one-to-one functions $g$ and $h$ are defined as follows.
g = \{(-9, 5), (0, 7), (3, -9), (6, 3), (8, 6)\}
$h(x) = 4x - 9$
Find the following.
g^{-1}(-9) =
h^{-1}(x) =
(h \circ h^{-1})(-5) =
1. **State the problem:**
We are given two one-to-one functions:
- $g = \{(-9, 5), (0, 7), (3, -9), (6, 3), (8, 6)\}$
- $h(x) = 4x - 9$
We need to find:
- $g^{-1}(-9)$
- $h^{-1}(x)$
- $(h \circ h^{-1})(-5)$
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2. **Find $g^{-1}(-9)$:**
The inverse function $g^{-1}$ reverses the pairs of $g$. That means if $g(a) = b$, then $g^{-1}(b) = a$.
From $g$, find the pair where the output is $-9$:
$$g(3) = -9$$
Therefore,
$$g^{-1}(-9) = 3$$
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3. **Find $h^{-1}(x)$:**
Given $h(x) = 4x - 9$, to find the inverse function $h^{-1}(x)$, we solve for $x$ in terms of $y$:
$$y = 4x - 9$$
Add $9$ to both sides:
$$y + 9 = 4x$$
Divide both sides by $4$:
$$\frac{y + 9}{4} = x$$
Using cancellation notation:
$$x = \frac{\cancel{4} \cdot \frac{y + 9}{\cancel{4}}}{1} = \frac{y + 9}{4}$$
So,
$$h^{-1}(x) = \frac{x + 9}{4}$$
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4. **Find $(h \circ h^{-1})(-5)$:**
By definition, $(h \circ h^{-1})(x) = h(h^{-1}(x))$.
Substitute $x = -5$:
$$h(h^{-1}(-5))$$
First find $h^{-1}(-5)$:
$$h^{-1}(-5) = \frac{-5 + 9}{4} = \frac{4}{4} = 1$$
Now find $h(1)$:
$$h(1) = 4(1) - 9 = 4 - 9 = -5$$
Therefore,
$$(h \circ h^{-1})(-5) = -5$$
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**Final answers:**
- $g^{-1}(-9) = 3$
- $h^{-1}(x) = \frac{x + 9}{4}$
- $(h \circ h^{-1})(-5) = -5$