How do you use composition of functions to show that #f(x)=(2+x)/x# and #f^1(x) = 2/(x1)# are inverses?
To show that two functions, ( f(x) = \frac{2+x}{x} ) and ( f^{1}(x) = \frac{2}{x1} ), are inverses of each other using composition of functions, we need to demonstrate that ( f(f^{1}(x)) = x ) and ( f^{1}(f(x)) = x ).

First, let's find ( f(f^{1}(x)) ): [ f(f^{1}(x)) = f\left(\frac{2}{x1}\right) ] Substitute ( \frac{2}{x1} ) into ( f(x) ): [ f\left(\frac{2}{x1}\right) = \frac{2 + \frac{2}{x1}}{\frac{2}{x1}} ] Simplify: [ \frac{2 + \frac{2}{x1}}{\frac{2}{x1}} = \frac{2(x1) + 2}{2} ] [ = \frac{2x  2 + 2}{2} = \frac{2x}{2} = x ]

Now, let's find ( f^{1}(f(x)) ): [ f^{1}(f(x)) = f^{1}\left(\frac{2+x}{x}\right) ] Substitute ( \frac{2+x}{x} ) into ( f^{1}(x) ): [ f^{1}\left(\frac{2+x}{x}\right) = \frac{2}{\frac{2+x}{x}1} ] [ = \frac{2}{\frac{2+xx}{x}} = \frac{2}{\frac{2}{x}} ] [ = \frac{2x}{2} = x ]
Since ( f(f^{1}(x)) = x ) and ( f^{1}(f(x)) = x ), the functions ( f(x) ) and ( f^{1}(x) ) are inverses of each other.
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Step by step explanation is given below.
Thus were are able to show that the functions are inverses of each other.
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When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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