Answer :
To determine if the function [tex]\( f(x) = \frac{x - 7}{x + 4} \)[/tex] has an inverse, we need to check if [tex]\( f \)[/tex] is one-to-one. A function is one-to-one if each value of the function corresponds to exactly one input value.
To find the inverse of [tex]\( f(x) \)[/tex], we follow these steps:
1. Rewrite the function:
[tex]\[ y = \frac{x - 7}{x + 4} \][/tex]
2. Solve for [tex]\( x \)[/tex] in terms of [tex]\( y \)[/tex]:
[tex]\[ y = \frac{x - 7}{x + 4} \][/tex]
Multiply both sides by [tex]\( x + 4 \)[/tex]:
[tex]\[ y(x + 4) = x - 7 \][/tex]
Expand and rearrange to isolate [tex]\( x \)[/tex]:
[tex]\[ yx + 4y = x - 7 \][/tex]
Bring all terms involving [tex]\( x \)[/tex] to one side:
[tex]\[ yx - x = -4y - 7 \][/tex]
Factor out [tex]\( x \)[/tex]:
[tex]\[ x(y - 1) = -4y - 7 \][/tex]
Solve for [tex]\( x \)[/tex]:
[tex]\[ x = \frac{-4y - 7}{y - 1} \][/tex]
3. Substitute back to find [tex]\( f^{-1}(y) \)[/tex]:
This gives us:
[tex]\[ f^{-1}(x) = \frac{-4x - 7}{x - 1} \][/tex]
Simplify by multiplying numerator and denominator by -1:
[tex]\[ f^{-1}(x) = \frac{4x + 7}{1 - x} \][/tex]
4. Check the domain restriction:
In the original function [tex]\( f(x) = \frac{x - 7}{x + 4} \)[/tex], the denominator [tex]\( x + 4 \)[/tex] should not be zero. Therefore, [tex]\( x \neq -4 \)[/tex].
For the inverse function [tex]\( f^{-1}(x) = \frac{4x + 7}{1 - x} \)[/tex], the denominator [tex]\( 1 - x \)[/tex] should not be zero. Therefore, [tex]\( x \neq 1 \)[/tex].
5. Verification:
We should verify that both proposed inverse functions correctly invert the original function.
Check if [tex]\( f(f^{-1}(x)) = x \)[/tex] and [tex]\( f^{-1}(f(x)) = x \)[/tex]:
[tex]\[ f\left(f^{-1}(x)\right) = f\left(\frac{4x + 7}{1 - x}\right) = \frac{\left(\frac{4x + 7}{1 - x}\right) - 7}{\left(\frac{4x + 7}{1 - x}\right) + 4} \][/tex]
Simplify:
[tex]\[ f\left(\frac{4x + 7}{1 - x}\right) = \frac{\frac{4x + 7 - 7(1 - x)}{1 - x}}{\frac{4x + 7 + 4(1 - x)}{1 - x}} = \frac{\frac{4x + 7 - 7 + 7x}{1 - x}}{\frac{4x + 7 + 4 - 4x}{1 - x}} = \frac{\frac{11x}{1 - x}}{\frac{11}{1 - x}} = x \][/tex]
And:
[tex]\[ f^{-1}\left(f(x)\right) = f^{-1}\left(\frac{x - 7}{x + 4}\right) = \frac{4\left(\frac{x - 7}{x + 4}\right) + 7}{1 - \left(\frac{x - 7}{x + 4}\right)} \][/tex]
Simplify:
[tex]\[ f^{-1}\left(\frac{x - 7}{x + 4}\right) = \frac{\frac{4(x - 7) + 7(x + 4)}{x + 4}}{\frac{x + 4 - (x - 7)}{x + 4}} = \frac{\frac{4x - 28 + 7x + 28}{x + 4}}{\frac{x + 4 - x + 7}{x + 4}} = \frac{\frac{11x}{x + 4}}{\frac{11}{x + 4}} = x \][/tex]
Since [tex]\( f\left(f^{-1}(x)\right) = x \)[/tex] and [tex]\( f^{-1}\left(f(x)\right) = x \)[/tex], we have correctly found:
[tex]\[ f^{-1}(x) = \frac{4x + 7}{1 - x}, \quad x \neq 1 \][/tex]
Therefore, the inverse function is [tex]\( f^{-1}(x) = \frac{4x + 7}{1 - x} \)[/tex] with the restriction that [tex]\( x \neq 1 \)[/tex].
To find the inverse of [tex]\( f(x) \)[/tex], we follow these steps:
1. Rewrite the function:
[tex]\[ y = \frac{x - 7}{x + 4} \][/tex]
2. Solve for [tex]\( x \)[/tex] in terms of [tex]\( y \)[/tex]:
[tex]\[ y = \frac{x - 7}{x + 4} \][/tex]
Multiply both sides by [tex]\( x + 4 \)[/tex]:
[tex]\[ y(x + 4) = x - 7 \][/tex]
Expand and rearrange to isolate [tex]\( x \)[/tex]:
[tex]\[ yx + 4y = x - 7 \][/tex]
Bring all terms involving [tex]\( x \)[/tex] to one side:
[tex]\[ yx - x = -4y - 7 \][/tex]
Factor out [tex]\( x \)[/tex]:
[tex]\[ x(y - 1) = -4y - 7 \][/tex]
Solve for [tex]\( x \)[/tex]:
[tex]\[ x = \frac{-4y - 7}{y - 1} \][/tex]
3. Substitute back to find [tex]\( f^{-1}(y) \)[/tex]:
This gives us:
[tex]\[ f^{-1}(x) = \frac{-4x - 7}{x - 1} \][/tex]
Simplify by multiplying numerator and denominator by -1:
[tex]\[ f^{-1}(x) = \frac{4x + 7}{1 - x} \][/tex]
4. Check the domain restriction:
In the original function [tex]\( f(x) = \frac{x - 7}{x + 4} \)[/tex], the denominator [tex]\( x + 4 \)[/tex] should not be zero. Therefore, [tex]\( x \neq -4 \)[/tex].
For the inverse function [tex]\( f^{-1}(x) = \frac{4x + 7}{1 - x} \)[/tex], the denominator [tex]\( 1 - x \)[/tex] should not be zero. Therefore, [tex]\( x \neq 1 \)[/tex].
5. Verification:
We should verify that both proposed inverse functions correctly invert the original function.
Check if [tex]\( f(f^{-1}(x)) = x \)[/tex] and [tex]\( f^{-1}(f(x)) = x \)[/tex]:
[tex]\[ f\left(f^{-1}(x)\right) = f\left(\frac{4x + 7}{1 - x}\right) = \frac{\left(\frac{4x + 7}{1 - x}\right) - 7}{\left(\frac{4x + 7}{1 - x}\right) + 4} \][/tex]
Simplify:
[tex]\[ f\left(\frac{4x + 7}{1 - x}\right) = \frac{\frac{4x + 7 - 7(1 - x)}{1 - x}}{\frac{4x + 7 + 4(1 - x)}{1 - x}} = \frac{\frac{4x + 7 - 7 + 7x}{1 - x}}{\frac{4x + 7 + 4 - 4x}{1 - x}} = \frac{\frac{11x}{1 - x}}{\frac{11}{1 - x}} = x \][/tex]
And:
[tex]\[ f^{-1}\left(f(x)\right) = f^{-1}\left(\frac{x - 7}{x + 4}\right) = \frac{4\left(\frac{x - 7}{x + 4}\right) + 7}{1 - \left(\frac{x - 7}{x + 4}\right)} \][/tex]
Simplify:
[tex]\[ f^{-1}\left(\frac{x - 7}{x + 4}\right) = \frac{\frac{4(x - 7) + 7(x + 4)}{x + 4}}{\frac{x + 4 - (x - 7)}{x + 4}} = \frac{\frac{4x - 28 + 7x + 28}{x + 4}}{\frac{x + 4 - x + 7}{x + 4}} = \frac{\frac{11x}{x + 4}}{\frac{11}{x + 4}} = x \][/tex]
Since [tex]\( f\left(f^{-1}(x)\right) = x \)[/tex] and [tex]\( f^{-1}\left(f(x)\right) = x \)[/tex], we have correctly found:
[tex]\[ f^{-1}(x) = \frac{4x + 7}{1 - x}, \quad x \neq 1 \][/tex]
Therefore, the inverse function is [tex]\( f^{-1}(x) = \frac{4x + 7}{1 - x} \)[/tex] with the restriction that [tex]\( x \neq 1 \)[/tex].
