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What is the inverse function of #f(x)=x+1# and how do you find #f^-1(2)#?

Answer 1

Serenity Jones

#f^(-1)(x) = x-1#

#f^(-1)(2) = 1#

#f^(-1)(x)# is a function such that #f(f^(-1))(x) = f^(-1)(f(x)) = x#

Then, if #y = f^(-1)(x)# we can apply #f(x)# to both sides to get

#f(y) = f(f^(-1)(x)) = x#

In the given function, this translates to

#y + 1 = x#

Then, as #y = f^(-1)(x)#, we can simply solve for #y# and substitute.

#y = x-1#

#=> f^(-1)(x) = x-1#

And so #f^(-1)(2) = 2-1 = 1#

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Answer 2

Caleb Alexander

The inverse function of ( f(x) = x + 1 ) is ( f^{-1}(x) = x - 1 ). To find ( f^{-1}(2) ), substitute ( x = 2 ) into the inverse function, which gives ( f^{-1}(2) = 2 - 1 = 1 ). Therefore, ( f^{-1}(2) = 1 ).

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Answer from HIX Tutor

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.