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To find the inverse function [tex]\( h^{-1}(x) \)[/tex] and determine the domain and range of [tex]\( h^{-1}(x) \)[/tex], let's proceed with the following steps.
### Step 1: Find the Inverse Function [tex]\( h^{-1}(x) \)[/tex]
Given the function [tex]\( h(x) = \frac{7x - 3}{5x + 9} \)[/tex], we want to find its inverse [tex]\( h^{-1}(x) \)[/tex].
1. Rewrite [tex]\( h(x) \)[/tex] as [tex]\( y \)[/tex]:
[tex]\[ y = \frac{7x - 3}{5x + 9} \][/tex]
2. Solve for [tex]\( x \)[/tex] in terms of [tex]\( y \)[/tex]: Start by swapping [tex]\( x \)[/tex] and [tex]\( y \)[/tex] to find the inverse:
[tex]\[ x = \frac{7y - 3}{5y + 9} \][/tex]
3. Cross-multiply to solve for [tex]\( y \)[/tex]:
[tex]\[ x(5y + 9) = 7y - 3 \][/tex]
Expand and combine like terms:
[tex]\[ 5xy + 9x = 7y - 3 \][/tex]
[tex]\[ 5xy - 7y = -9x - 3 \][/tex]
[tex]\[ y(5x - 7) = -9x - 3 \][/tex]
4. Isolate [tex]\( y \)[/tex]:
[tex]\[ y = \frac{-9x - 3}{5x - 7} \][/tex]
So, the inverse function is:
[tex]\[ h^{-1}(x) = \frac{-9x - 3}{5x - 7} \][/tex]
### Step 2: Determine the Domain of [tex]\( h^{-1}(x) \)[/tex]
The domain of [tex]\( h^{-1}(x) \)[/tex] is the range of [tex]\( h(x) \)[/tex]. To find the range of [tex]\( h(x) \)[/tex], we need to consider the values that [tex]\( y = \frac{7x - 3}{5x + 9} \)[/tex] can take.
1. Identify any restrictions on [tex]\( y \)[/tex]:
The function [tex]\( h(x) = \frac{7x - 3}{5x + 9} \)[/tex] has a vertical asymptote where the denominator is zero:
[tex]\[ 5x + 9 = 0 \implies x = -\frac{9}{5} \][/tex]
For all other [tex]\( x \)[/tex], the function [tex]\( h(x) \)[/tex] is defined.
2. Check if there are any values that [tex]\( y \)[/tex] cannot take by solving for [tex]\( x \)[/tex] in terms of [tex]\( y \)[/tex]:
From [tex]\( y = \frac{7x - 3}{5x + 9} \)[/tex], as [tex]\( x \)[/tex] approaches [tex]\(-\frac{9}{5}\)[/tex], [tex]\( y \)[/tex] approaches a vertical asymptote. But since the function is a rational function of degree 1, all real numbers can be achieved by [tex]\( y \)[/tex].
Therefore, the range of [tex]\( h(x) \)[/tex], and thus the domain of [tex]\( h^{-1}(x) \)[/tex], is all real numbers:
[tex]\[ \text{Domain of } h^{-1}(x): (-\infty, \infty) \][/tex]
### Step 3: Determine the Range of [tex]\( h^{-1}(x) \)[/tex]
The range of [tex]\( h^{-1}(x) \)[/tex] is the domain of [tex]\( h(x) \)[/tex].
As established:
[tex]\[ h(x) = \frac{7x - 3}{5x + 9} \][/tex]
The domain of [tex]\( h(x) \)[/tex] excludes the value where the denominator is zero:
[tex]\[ 5x + 9 = 0 \implies x = -\frac{9}{5} \][/tex]
Hence, the domain of [tex]\( h(x) \)[/tex] and therefore the range of [tex]\( h^{-1}(x) \)[/tex] is:
[tex]\[ \text{Range of } h^{-1}(x): (-\infty, -\frac{9}{5}) \cup (-\frac{9}{5}, \infty) \][/tex]
### Summary:
- Inverse function: [tex]\[ h^{-1}(x) = \frac{-9x - 3}{5x - 7} \][/tex]
- Domain of [tex]\( h^{-1}(x) \)[/tex]: [tex]\((-\infty, \infty)\)[/tex]
- Range of [tex]\( h^{-1}(x) \)[/tex]: [tex]\((-\infty, -\frac{9}{5}) \cup (-\frac{9}{5}, \infty)\)[/tex]
### Step 1: Find the Inverse Function [tex]\( h^{-1}(x) \)[/tex]
Given the function [tex]\( h(x) = \frac{7x - 3}{5x + 9} \)[/tex], we want to find its inverse [tex]\( h^{-1}(x) \)[/tex].
1. Rewrite [tex]\( h(x) \)[/tex] as [tex]\( y \)[/tex]:
[tex]\[ y = \frac{7x - 3}{5x + 9} \][/tex]
2. Solve for [tex]\( x \)[/tex] in terms of [tex]\( y \)[/tex]: Start by swapping [tex]\( x \)[/tex] and [tex]\( y \)[/tex] to find the inverse:
[tex]\[ x = \frac{7y - 3}{5y + 9} \][/tex]
3. Cross-multiply to solve for [tex]\( y \)[/tex]:
[tex]\[ x(5y + 9) = 7y - 3 \][/tex]
Expand and combine like terms:
[tex]\[ 5xy + 9x = 7y - 3 \][/tex]
[tex]\[ 5xy - 7y = -9x - 3 \][/tex]
[tex]\[ y(5x - 7) = -9x - 3 \][/tex]
4. Isolate [tex]\( y \)[/tex]:
[tex]\[ y = \frac{-9x - 3}{5x - 7} \][/tex]
So, the inverse function is:
[tex]\[ h^{-1}(x) = \frac{-9x - 3}{5x - 7} \][/tex]
### Step 2: Determine the Domain of [tex]\( h^{-1}(x) \)[/tex]
The domain of [tex]\( h^{-1}(x) \)[/tex] is the range of [tex]\( h(x) \)[/tex]. To find the range of [tex]\( h(x) \)[/tex], we need to consider the values that [tex]\( y = \frac{7x - 3}{5x + 9} \)[/tex] can take.
1. Identify any restrictions on [tex]\( y \)[/tex]:
The function [tex]\( h(x) = \frac{7x - 3}{5x + 9} \)[/tex] has a vertical asymptote where the denominator is zero:
[tex]\[ 5x + 9 = 0 \implies x = -\frac{9}{5} \][/tex]
For all other [tex]\( x \)[/tex], the function [tex]\( h(x) \)[/tex] is defined.
2. Check if there are any values that [tex]\( y \)[/tex] cannot take by solving for [tex]\( x \)[/tex] in terms of [tex]\( y \)[/tex]:
From [tex]\( y = \frac{7x - 3}{5x + 9} \)[/tex], as [tex]\( x \)[/tex] approaches [tex]\(-\frac{9}{5}\)[/tex], [tex]\( y \)[/tex] approaches a vertical asymptote. But since the function is a rational function of degree 1, all real numbers can be achieved by [tex]\( y \)[/tex].
Therefore, the range of [tex]\( h(x) \)[/tex], and thus the domain of [tex]\( h^{-1}(x) \)[/tex], is all real numbers:
[tex]\[ \text{Domain of } h^{-1}(x): (-\infty, \infty) \][/tex]
### Step 3: Determine the Range of [tex]\( h^{-1}(x) \)[/tex]
The range of [tex]\( h^{-1}(x) \)[/tex] is the domain of [tex]\( h(x) \)[/tex].
As established:
[tex]\[ h(x) = \frac{7x - 3}{5x + 9} \][/tex]
The domain of [tex]\( h(x) \)[/tex] excludes the value where the denominator is zero:
[tex]\[ 5x + 9 = 0 \implies x = -\frac{9}{5} \][/tex]
Hence, the domain of [tex]\( h(x) \)[/tex] and therefore the range of [tex]\( h^{-1}(x) \)[/tex] is:
[tex]\[ \text{Range of } h^{-1}(x): (-\infty, -\frac{9}{5}) \cup (-\frac{9}{5}, \infty) \][/tex]
### Summary:
- Inverse function: [tex]\[ h^{-1}(x) = \frac{-9x - 3}{5x - 7} \][/tex]
- Domain of [tex]\( h^{-1}(x) \)[/tex]: [tex]\((-\infty, \infty)\)[/tex]
- Range of [tex]\( h^{-1}(x) \)[/tex]: [tex]\((-\infty, -\frac{9}{5}) \cup (-\frac{9}{5}, \infty)\)[/tex]
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