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Characterize the inverse function of function $f$, so that $f$ is defined by:

a. $f(x) = \log_2{(x+1)}$, with:

  1. $D_f = ]-1,+\infty[$
  2. $D_f = ]-1,7]$

b. $f(x) = 2-\ln{(\frac{x}{3})}$, with:

  1. $D_f = ]0,+\infty[$
  2. $D_f = ]0,3e]$

I have already characterized the functions and verified them:

a.$f^{-1}(x) = 2^x-1$

b.$f^{-1}(x) = 3e^{2-x}$

My question is that I am not sure of what the second part of this problem means. According to my book, the solution is:

a.

  1. $f^1(x) = 2^x-1; D_{f^{-1}} = \mathbb{R}$
  2. $f^{-1}(x)=2^x-1; D_{f^{-1}}=]-\infty,3]$

b.

  1. $f^{-1}(x) = 3e^{2-x}; D_{f^{-1}} = \mathbb{R}$

  2. $f^{-1}(x) = 3e^{2-x}; D_{f^{-1}} = [1,+\infty[$

What does this mean exactly? My guess is that I am supposed to find the possible input/domain of $f^{-1}$ inside the given input for $f$ ?

Can anyone explain this to me?

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1 Answer 1

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You have to think about the images of $D_f$, because we know that $f:X\to Y$ is inversible if $f$ is bijective, that is, if $f$ is injective and $f(X)=Y$. Then $f^{-1}$ is defined from $Y$ to $X$ without 'problems'.

For example, if $f(x)=\log_2(x+1)$ and $D_f=[0,3]$, then $f(D_f)=[0,2]$ (why?) and therefore $f^{-1}(x)=2^x-1$ is defined on $[0,2]$.

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