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I was given the function $f(x) = \sqrt{2x+5}$ $x\in\Bbb R$ , $x \geq -2.5$

I was asked to find the inverse function which was $f^{-1}(x) = {x^2-5\over 2}$

Then I was asked to find the range of $f(x)$ and the domain of $f^{-1}(x)$ which is $x \geq 0$ and $f^{-1}(x) \geq 0$. This is because the range and domain swap for a function and its inverse. However if a value less than $0$ is in-putted for $x$ in the inverse function it obviously still works due to the fact that the inverse function is a quadratic.

Would someone be able to explain to me why the domain of the inverse function is not $x\in\Bbb R$?

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  • $\begingroup$ The symbol ‘belongs to’ is obtained with the command \in. ‘>=’ , with \ge or \geq. $\endgroup$
    – Bernard
    Commented Sep 21, 2015 at 15:12

3 Answers 3

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You wrote it yourself: it is because the domain of $f^{-1}$ is the range of $f$. Don't forget defining a function consists in defining, not only a computational formula, but its domain of validity, and its target.

So, although $\dfrac{x^2-5}2$ is a perfectly valid function defining formula for all $x\in\mathbf R$, it does not define, per se, the inverse of $f$. If you consider it defines a function with domain $\mathbf R$, this function is a continuation of $f^{-1}$, which is defined on $\mathbf R^+$, to the whole of $\mathbf R$.

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  • $\begingroup$ You meant the inverse of $f$ rather than its reciprocal. $\endgroup$ Commented Sep 21, 2015 at 17:04
  • $\begingroup$ Oh yes! I'm not a native English speaker, and in French, the terms employed are exchanged w.r.t. English. I've corrected the post. Thanks for pointing it. $\endgroup$
    – Bernard
    Commented Sep 21, 2015 at 18:36
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Hint: By definition $f(f^{-1}(x))=f^{-1}(f(x))$ and use domain of composition of functions.

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Take care that $f^{-1}$ is not only the formal expression $$f^{-1}(x)=\frac{x^2-5}{2}$$ Since $f^{-1}$ is the function that take some element $y$ of $\text{Range }(f)$ and map it to one element $x$ of $\text{Dom }(f)$, such that $y=f(x)$. So, for $x<0$, the expression $f^{-1}$ has no sense since there is no element lees than $0$ in $\text{Range }(f)$.

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