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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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