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if we have the following functions:

$f(x)=x^3+x$

$g(x)=x^3+x-9$

$h(x)=x|x|+1$

is there a simple way to calculate these inverses:

$f^{-1}(2)$

$g^{-1}(1)$

$h^{-1}(3)$

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closed as off-topic by CY Aries, GNUSupporter 8964民主女神 地下教會, Namaste, Ethan Bolker, Jose Arnaldo Bebita-Dris Mar 31 '18 at 12:40

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  • $\begingroup$ What do you mean by simple? Solving $x^3+x=x(x^2+1)=2$ should be quite doable isn't it? (In particular if you note that $1^2 = 1$). $\endgroup$ – Surb Mar 31 '18 at 8:54
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Note that since $f(x)=x^3+x$ is injective since $f'(x)=x^2+1>0$

  • $f(x)=x^3+x=2\implies x=1 \implies f^{-1}(2)=1$

and also for $g(x)=x^3+x-9$ injective since $g'(x)=x^2+1>0$

  • $g(x)=x^3+x-9=1\implies x=2 \implies g^{-1}(1)=2$

and also for $h(x)=x|x|+1$ injective (can you say why?) we can find

  • $h(x)=x|x|+1=3\implies x=\sqrt 2 \implies h^{-1}(3)=\sqrt2$
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  • $\begingroup$ thank you, this is what i did i just wanted to make sure if it's correct or not $\endgroup$ – SiN Mar 31 '18 at 9:15
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    $\begingroup$ @SiN You are welcome. For next times show also your effort and work in the OP in order to better help you with your doubts and also to avoid downvoting for a poor format. Bye $\endgroup$ – gimusi Mar 31 '18 at 9:17
  • $\begingroup$ @SiN The key point here is injectivity,once you show that the functions are injective you are sure that the solution we can find by inspection in unique and then we can find the values for the inverses. $\endgroup$ – gimusi Mar 31 '18 at 9:19
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For $$h(x)=x|x|+1$$ it is not hard to find the inverse function, we only need $h^{-1}(x)$ for $x\geq 0$, so we have $$h^{-1}(x)=\sqrt{x-1}$$

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