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This question already has an answer here:

  • Show that the image of the function $f:(0,\infty)\rightarrow \mathbb{R}$, $f(x)=x+\dfrac{1}{x}$ is the interval $[2,\infty)$.

If $x=1$, then $f(1)=2$. So how can I show that the mage of the function is the interval $[2,\infty)$?

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marked as duplicate by José Carlos Santos, Jyrki Lahtonen, Asaf Karagila elementary-set-theory Apr 21 at 18:20

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ Solve the equation $x+\frac 1x=y$ whenever $y\ge2$? $\endgroup$ – Lord Shark the Unknown Apr 21 at 18:09
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    $\begingroup$ Check if it is increasing decreasing in general extreme values $\endgroup$ – AmerYR Apr 21 at 18:10
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    $\begingroup$ Hint: use AM-GM inequality or simply complete the square $\endgroup$ – Μάρκος Καραμέρης Apr 21 at 18:11
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    $\begingroup$ @Jyrki Lahtonen Clearly, it is not the same $\endgroup$ – Aqua Apr 21 at 18:23
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    $\begingroup$ @José Carlos Santos Clearly, it is not the same $\endgroup$ – Aqua Apr 21 at 18:25
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Since for positive $x$ by Am-Gm we have $$x+{1\over x}\geq 2$$ with equality iff $x=1$ we see that the range is $[2,\infty)$ since $f$ is continuous on $(0,\infty)$. Note that here is continuation essentialy, without it the claim is not necessary true.


Alternatively, you can avoid continuity. You need to find out for which $y$ is $$x+{1\over x}=y$$ solvable, so when is $x^2-yx+1=0$ solvable. That is true iff it discriminat is nonegative, i.e. $$y^2-4\geq 0\implies (y-2)(y+2)\geq 0$$ Since clearly $y=x+{1\over x}>0$ we have $y-2\geq 0$ so the range is $[2,\infty)$.

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  • $\begingroup$ Thanks for answer and comments. $\endgroup$ – PozcuKushimotoStreet Apr 21 at 19:04
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Hint: It is $$x+\frac{1}{x}\geq 2$$ and this is equivalent to $$(x-1)^2\geq 0$$ if $$x>0$$

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