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if f(x)=2x/x-5 find f^-1(x)

Sagot :

Answer:

[tex]f^{-1}[/tex] (x) = [tex]\frac{5x}{x-2}[/tex]

Step-by-step explanation:

let y = f(x) and rearrange making x the subject

y = [tex]\frac{2x}{x-5}[/tex] ( multiply both sides by x - 5 )

y(x - 5) = 2x ← distribute left side

xy - 5y = 2x ( subtract 2x from both sides )

xy - 2x - 5y = 0 ( add 5y to both sides )

xy - 2x = 5y ← factor out x from each term on the left side )

x(y - 2) - 5y ← divide both sides by y - 2

x = [tex]\frac{5y}{y-2}[/tex]

Change y back into terms of x with x = [tex]f^{-1}[/tex] (x) , then

[tex]f^{-1}[/tex] (x) = [tex]\frac{5x}{x-2}[/tex]

Answer:

Look into the image. I hope it helps❤

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