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Arrange the equations in the correct sequence to find the inverse of [tex]$f(x)=y=\frac{3 x}{8+x}$[/tex].

[tex]\[
\begin{array}{c}
x=\frac{3 y}{8+y} \\
x(8+y)=3 y \\
8 x=3 y-x y \\
8 x=y(3-x) \\
y=f^{-1}(x)=\frac{8 x}{3-x} \\
\end{array}
\][/tex]


Sagot :

To find the inverse [tex]\( f^{-1}(x) \)[/tex] of the function [tex]\( f(x)=y=\frac{3x}{8+x} \)[/tex], we need to follow a series of algebraic steps. Let's arrange the given equations in the correct sequence:

1. Start with the given function in terms of [tex]\( y \)[/tex] and [tex]\( x \)[/tex]:

[tex]\[ x = \frac{3y}{8 + y} \][/tex]

2. Multiply both sides by [tex]\( 8 + y \)[/tex] to clear the denominator:

[tex]\[ x(8 + y) = 3y \][/tex]

3. Distribute [tex]\( x \)[/tex] on the left side:

[tex]\[ 8x + xy = 3y \][/tex]

4. Move all terms involving [tex]\( y \)[/tex] to one side of the equation:

[tex]\[ 8x = 3y - xy \][/tex]

5. Factor [tex]\( y \)[/tex] on the right side:

[tex]\[ 8x = y(3 - x) \][/tex]

6. Solve for [tex]\( y \)[/tex] by dividing both sides by [tex]\( 3 - x \)[/tex]:

[tex]\[ y = \frac{8x}{3 - x} \][/tex]

Hence, the inverse function is:

[tex]\[ y = f^{-1}(x) = \frac{8x}{3 - x} \][/tex]

So the correct sequence of equations is:

1. [tex]\( x = \frac{3y}{8 + y} \)[/tex]
2. [tex]\( x(8 + y) = 3y \)[/tex]
3. [tex]\( 8x + xy = 3y \)[/tex]
4. [tex]\( 8x = 3y - xy \)[/tex]
5. [tex]\( 8x = y(3 - x) \)[/tex]
6. [tex]\( y = f^{-1}(x) = \frac{8x}{3 - x} \)[/tex]