# If #f(x+1) =2x# and #g(3x) = x+6#, how do you find the value of #f^-1(g(f(13)))3?

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First, let's break down the expression step by step:

- Start with the innermost function, ( f(13) ), which means substituting ( x = 13 ) into ( f(x+1) = 2x ):

[ f(13+1) = 2 \cdot 13 ] [ f(14) = 26 ]

- Next, apply the function ( g(x) = x + 6 ) to the result from step 1, ( g(f(14)) ):

[ g(f(14)) = f(14) + 6 ] [ g(f(14)) = 26 + 6 ] [ g(f(14)) = 32 ]

- Now, apply the inverse of ( f(x+1) ) to the result from step 2, ( f^{-1}(g(f(14))) ). Since ( f(x+1) = 2x ), the inverse function ( f^{-1}(y) ) undoes the operation of doubling and subtracting 1:

[ f^{-1}(32) = \frac{32}{2} - 1 ] [ f^{-1}(32) = 16 - 1 ] [ f^{-1}(32) = 15 ]

- Finally, raise the result from step 3 to the power of 3:

[ f^{-1}(g(f(13)))^3 = 15^3 ] [ f^{-1}(g(f(13)))^3 = 3375 ]

Therefore, the value of ( f^{-1}(g(f(13)))^3 ) is 3375.

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When evaluating a one-sided 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 one-sided 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 one-sided 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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