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For which pairs of functions is (f•g)(x) = x?

fx)=x^2 and g(x)= 1/x
f{x) = 2/x and g(x) = 2/x
f(x)= x-2/3 and g(x) = 2-3/x
f(x) = 1/2x - 2 and g(x) = 1/2x+ 2

Sagot :

 Rules for the operations of the functions state,

  • (f o g)(x) = f[g(x)]

Option (2) is the correct option.

    Now we will apply this operation in each option to check (f o g)(x) = x

Option (1)

f(x) = x² and g(x) = [tex]\frac{1}{x}[/tex]

Therefore, (f o g)(x) = f[g(x)]

                                = [tex]\frac{1}{x^2}[/tex]

False.

Option (2)

f(x) = [tex]\frac{2}{x}[/tex] and g(x) =

(f o g)(x) = [tex]\frac{2}{\frac{2}{x}}[/tex]

             = [tex]\frac{2}{1}\times \frac{x}{2}[/tex]

             = x

True.

Option (3)

f(x) = [tex]\frac{x-2}{3}[/tex] and [tex]g(x) = 2-3x[/tex]

(f o g)(x) = f[g(x)]

             = [tex]\frac{2-3x-2}{3}[/tex]

             = -x

False.

Option (4)

f(x) = [tex]\frac{1}{2x}-2[/tex] and [tex]g(x)=\frac{1}{2x}+2[/tex]

(f o g)(x) = f[g(x)

             = [tex]\frac{1}{2(\frac{1}{2x}+2)}-2[/tex]

             = [tex]\frac{1}{\frac{1}{x}+4}-2[/tex]

             = [tex]\frac{x}{1+4x}-2[/tex]

             = [tex]\frac{x-2(4x+1)}{4x+1}[/tex]

             = [tex]\frac{x-8x-2}{4x+1}[/tex]

             = [tex]-\frac{7x+2}{4x+1}[/tex]

False.

                Therefore, Option (2) is the correct option.

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