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$f(x) = x + \dfrac{k}{x} $ ($k > 0$ and $x > 0$)

given that the k is a constant, how do I solve the x that makes f(x) minimum?

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  • $\begingroup$ Is $x > 0$ here? And is $k > 0$? $\endgroup$ Jun 16, 2019 at 12:06
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    $\begingroup$ My guess is that it is minimized when $x=\sqrt{k}$. $\endgroup$
    – kingW3
    Jun 16, 2019 at 12:07

4 Answers 4

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If you don't like differentiating, by AM-GM

$$\sqrt{k}=\sqrt{x\cdot\dfrac{k}{x}}\leq\dfrac{x+\dfrac{k}{x}}{2}$$

When will you have equality?

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    $\begingroup$ This is a nice answer! Using AM-GM is a clever idea. $\endgroup$ Jun 16, 2019 at 12:11
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To avoid differentiating, notice that $$x+\frac{k}{x}=\bigg(\sqrt{x}-\sqrt{\frac{k}{x}}\bigg)^2+2\sqrt{k}$$ and that $\sqrt{x}=\sqrt{k/x}$ when $x=\sqrt{k}$.

(Note that we can use $\sqrt{x}$, since it is given that $x\gt 0$)

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    $\begingroup$ why avoid differentiating? $\endgroup$ Jun 16, 2019 at 12:18
  • $\begingroup$ @AdamRubinson Why use “advanced” methods when the problem can be solved by simply completing the square? There’s something beautiful about a simple solution... $\endgroup$ Jun 16, 2019 at 12:20
  • $\begingroup$ I understand the answer now. It's a neat solution. $\endgroup$ Jun 16, 2019 at 12:36
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$$x+\frac{k}{x}=\frac{1}{x}\Big[(x-\sqrt k)^2+2\sqrt k x\Big]$$ $$=2\sqrt k+\frac{(x-\sqrt k)^2}{x}\geq2\sqrt k$$ Takes its minimum at $x=\sqrt k$

I have assumed $x>0$. Do the $x<0$ for yourself. Hint: This time the function has a maximum!

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If $x>0$ we have that $$x+\frac1x\geq 2\\$$ This follows from $$(x-1)^2\geq 0\\x^2-2x+1\geq 0\\x^2+1\geq 2x\\\frac{x^2+1}{x}\geq 2\\x+\frac1x\geq 2$$

Now if we put $x=t\sqrt{k}$ then we have $$t\sqrt{k}+\frac{\sqrt{k}}{t}=\sqrt{k}(t+\frac1t)\geq 2\sqrt{k}$$ The minimum is indeed achieved for $t=1$ or $x=\sqrt{k}$.

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