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Given f(x)= 1/1-x and g(x)= x-1/x
find
f of g
g of f^-1
g^-1​

Sagot :

semsee45

Answer:

[tex]f(g(x))=x[/tex]

[tex]g(f^{-1}(x))=\dfrac{-1}{x-1}[/tex]

[tex]g^{-1}(x)=\dfrac{1}{1-x}[/tex]

Step-by-step explanation:

[tex]f(x)=\dfrac{1}{1-x} \ \ \textsf{and} \ \ g(x)=\dfrac{x-1}{x}[/tex]

[tex]f(g(x))=\dfrac{1}{1-(\frac{x-1}{x})}[/tex]

[tex]\implies f(g(x))=\dfrac{1}{\frac{x}{x}-\frac{(x-1)}{x}}[/tex]

[tex]\implies f(g(x))=\dfrac{1}{\frac{x-x+1}{x}}[/tex]

[tex]\implies f(g(x))=\dfrac{1}{\frac{1}{x}}[/tex]

[tex]\implies f(g(x))=x[/tex]

[tex]x=\dfrac{1}{1-y}[/tex]

[tex]\implies 1-y=\dfrac{1}{x}[/tex]

[tex]\implies y=1-\dfrac{1}{x}[/tex]

[tex]\implies f^{-1}(x)=1-\dfrac{1}{x}[/tex]

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

[tex]\implies g(f^{-1}(x))=1-\dfrac{1}{\frac{x-1}{x}}[/tex]

[tex]\implies g(f^{-1}(x))=1-\dfrac{x}{x-1}[/tex]

[tex]\implies g(f^{-1}(x))=\dfrac{x-1-x}{x-1}[/tex]

[tex]\implies g(f^{-1}(x))=\dfrac{-1}{x-1}[/tex]

[tex]x=\dfrac{y-1}{y}[/tex]

[tex]\implies xy=y-1[/tex]

[tex]\implies 1=y-xy[/tex]

[tex]\implies 1=y(1-x)[/tex]

[tex]\implies y=\dfrac{1}{1-x}[/tex]

[tex]\implies g^{-1}(x)=\dfrac{1}{1-x}[/tex]

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