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Sagot :
To find the inverse of the function [tex]\( h(x) \)[/tex] and to determine the domain and range of the inverse function, follow these steps:
### Step 1: Express [tex]\( h(x) \)[/tex] in terms of [tex]\( y \)[/tex]
Given the function:
[tex]\[ h(x) = \frac{7x - 5}{6x + 7} \][/tex]
Replace [tex]\( h(x) \)[/tex] with [tex]\( y \)[/tex]:
[tex]\[ y = \frac{7x - 5}{6x + 7} \][/tex]
### Step 2: Swap [tex]\( x \)[/tex] and [tex]\( y \)[/tex] to find the inverse
Interchange [tex]\( x \)[/tex] and [tex]\( y \)[/tex]:
[tex]\[ x = \frac{7y - 5}{6y + 7} \][/tex]
### Step 3: Solve for [tex]\( y \)[/tex]
Solve the equation above for [tex]\( y \)[/tex] in terms of [tex]\( x \)[/tex]:
Multiply both sides by [tex]\( 6y + 7 \)[/tex] to get rid of the denominator:
[tex]\[ x(6y + 7) = 7y - 5 \][/tex]
Expand the left-hand side:
[tex]\[ 6xy + 7x = 7y - 5 \][/tex]
Bring all terms involving [tex]\( y \)[/tex] to one side and the constants to the other side:
[tex]\[ 6xy - 7y = -5 - 7x \][/tex]
Factor out [tex]\( y \)[/tex] on the left-hand side:
[tex]\[ y(6x - 7) = -5 - 7x \][/tex]
Divide both sides by [tex]\( 6x - 7 \)[/tex]:
[tex]\[ y = \frac{-5 - 7x}{6x - 7} \][/tex]
Hence, the inverse function [tex]\( h^{-1}(x) \)[/tex] is:
[tex]\[ h^{-1}(x) = \frac{-7x - 5}{6x - 7} \][/tex]
### Step 4: Determine the domain and range of [tex]\( h^{-1}(x) \)[/tex]
#### Domain and range of [tex]\( h(x) \)[/tex]:
- The domain of [tex]\( h(x) \)[/tex] is all real numbers except where the denominator is zero. Set the denominator of [tex]\( h(x) \)[/tex] to zero and solve for [tex]\( x \)[/tex]:
[tex]\[ 6x + 7 = 0 \implies x = -\frac{7}{6} \][/tex]
So, the domain of [tex]\( h(x) \)[/tex] is:
[tex]\[ (-\infty, -\frac{7}{6}) \cup (-\frac{7}{6}, \infty) \][/tex]
- The range of [tex]\( h(x) \)[/tex] matches the domain of [tex]\( h^{-1}(x) \)[/tex], which excludes the same value where [tex]\( 6y + 7 = 0 \)[/tex] (since [tex]\( h(x) \)[/tex] maps to all [tex]\( y \)[/tex] except the value that makes the denominator of the inverse function zero).
#### Domain of [tex]\( h^{-1}(x) \)[/tex]:
[tex]\[ (-\infty, -\frac{7}{6}) \cup (-\frac{7}{6}, \infty) \][/tex]
#### Range of [tex]\( h^{-1}(x) \)[/tex]:
[tex]\[ (-\infty, -\frac{7}{6}) \cup (-\frac{7}{6}, \infty) \][/tex]
### Final Answer:
The inverse function [tex]\( h^{-1}(x) \)[/tex] is:
[tex]\[ h^{-1}(x) = \frac{-7x - 5}{6x - 7} \][/tex]
The domain and range of [tex]\( h^{-1}(x) \)[/tex] are both:
[tex]\[ (-\infty, -\frac{7}{6}) \cup (-\frac{7}{6}, \infty) \][/tex]
### Step 1: Express [tex]\( h(x) \)[/tex] in terms of [tex]\( y \)[/tex]
Given the function:
[tex]\[ h(x) = \frac{7x - 5}{6x + 7} \][/tex]
Replace [tex]\( h(x) \)[/tex] with [tex]\( y \)[/tex]:
[tex]\[ y = \frac{7x - 5}{6x + 7} \][/tex]
### Step 2: Swap [tex]\( x \)[/tex] and [tex]\( y \)[/tex] to find the inverse
Interchange [tex]\( x \)[/tex] and [tex]\( y \)[/tex]:
[tex]\[ x = \frac{7y - 5}{6y + 7} \][/tex]
### Step 3: Solve for [tex]\( y \)[/tex]
Solve the equation above for [tex]\( y \)[/tex] in terms of [tex]\( x \)[/tex]:
Multiply both sides by [tex]\( 6y + 7 \)[/tex] to get rid of the denominator:
[tex]\[ x(6y + 7) = 7y - 5 \][/tex]
Expand the left-hand side:
[tex]\[ 6xy + 7x = 7y - 5 \][/tex]
Bring all terms involving [tex]\( y \)[/tex] to one side and the constants to the other side:
[tex]\[ 6xy - 7y = -5 - 7x \][/tex]
Factor out [tex]\( y \)[/tex] on the left-hand side:
[tex]\[ y(6x - 7) = -5 - 7x \][/tex]
Divide both sides by [tex]\( 6x - 7 \)[/tex]:
[tex]\[ y = \frac{-5 - 7x}{6x - 7} \][/tex]
Hence, the inverse function [tex]\( h^{-1}(x) \)[/tex] is:
[tex]\[ h^{-1}(x) = \frac{-7x - 5}{6x - 7} \][/tex]
### Step 4: Determine the domain and range of [tex]\( h^{-1}(x) \)[/tex]
#### Domain and range of [tex]\( h(x) \)[/tex]:
- The domain of [tex]\( h(x) \)[/tex] is all real numbers except where the denominator is zero. Set the denominator of [tex]\( h(x) \)[/tex] to zero and solve for [tex]\( x \)[/tex]:
[tex]\[ 6x + 7 = 0 \implies x = -\frac{7}{6} \][/tex]
So, the domain of [tex]\( h(x) \)[/tex] is:
[tex]\[ (-\infty, -\frac{7}{6}) \cup (-\frac{7}{6}, \infty) \][/tex]
- The range of [tex]\( h(x) \)[/tex] matches the domain of [tex]\( h^{-1}(x) \)[/tex], which excludes the same value where [tex]\( 6y + 7 = 0 \)[/tex] (since [tex]\( h(x) \)[/tex] maps to all [tex]\( y \)[/tex] except the value that makes the denominator of the inverse function zero).
#### Domain of [tex]\( h^{-1}(x) \)[/tex]:
[tex]\[ (-\infty, -\frac{7}{6}) \cup (-\frac{7}{6}, \infty) \][/tex]
#### Range of [tex]\( h^{-1}(x) \)[/tex]:
[tex]\[ (-\infty, -\frac{7}{6}) \cup (-\frac{7}{6}, \infty) \][/tex]
### Final Answer:
The inverse function [tex]\( h^{-1}(x) \)[/tex] is:
[tex]\[ h^{-1}(x) = \frac{-7x - 5}{6x - 7} \][/tex]
The domain and range of [tex]\( h^{-1}(x) \)[/tex] are both:
[tex]\[ (-\infty, -\frac{7}{6}) \cup (-\frac{7}{6}, \infty) \][/tex]
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