The real number x when added to its inverse gives the minimum value of the sum at x equal to what?
According to me it is 2 as $x +(1/x) $ is always equal to greater than 2. But the answer is given as 1.
The real number x when added to its inverse gives the minimum value of the sum at x equal to
0
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maxima-minima
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0Is it any real $x$ or $x>0$? Because $(x-1)^2\geq 0 \Rightarrow x+\frac{1}{x}\geq 2$ if $x>0$! – 2017-02-26
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0Won't the answer be $-\infty$? – 2018-12-14
3 Answers
1
The stationary point for $x + \frac1x$ is $x_0$ s.t. $f^\prime(x_0)=0$. So, $f^\prime(x) = 1-\frac1{x^2} = 0$. Then we have $x_0 = \pm 1$. One point $x_0 = -1$ gives the maximum $(-2)$ for $x < 0$ and the other gives minimum $(2)$ for $x > 0$. See the plot
2
The question asks you at what value of $x$ you have the minimum sum. Thus the answer is $x=1$ (derive $x+\frac1x$ and look for stationary points), which gives you a sum $s=1+\frac11=1+1=2$
2
Hint:
Please read the question carefully. It says The real number x when added to its inverse gives the minimum value of the sum at what value of x?.
Now answer the question. Hope it helps.
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0Sorry , I don't get it – 2017-02-26