# Show that the image of the function $f:(0,\infty)\rightarrow \mathbb{R}$, $f(x)=x+\frac1x$ is the interval $[2,\infty)$. [duplicate]

• 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 $$[2,\infty)$$?

## marked as duplicate by José Carlos Santos, Jyrki Lahtonen, Asaf Karagila♦ elementary-set-theory StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Apr 21 at 18:20

• Solve the equation $x+\frac 1x=y$ whenever $y\ge2$? – Lord Shark the Unknown Apr 21 at 18:09
• Check if it is increasing decreasing in general extreme values – AmerYR Apr 21 at 18:10
• Hint: use AM-GM inequality or simply complete the square – Μάρκος Καραμέρης Apr 21 at 18:11
• @Jyrki Lahtonen Clearly, it is not the same – Aqua Apr 21 at 18:23
• @José Carlos Santos Clearly, it is not the same – Aqua Apr 21 at 18:25

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 $$[2,\infty)$$.
Hint: It is $$x+\frac{1}{x}\geq 2$$ and this is equivalent to $$(x-1)^2\geq 0$$ if $$x>0$$