This question related to maxima and minima concept so we have to use first and second derivative test.

first we have to asume that number so,

let x is required number

according the condition x + 1/x=as small as possible

now we make one function of x and it's reciprocal

f(x)=x+1/x

take first derivative of f(x)

f '(x)=1-1/x^2

put f '(x)=0 to get maximum and minimum value

1-1/x^2=0

x=+-1

so one is maximum value and one is minimum value, to find which one is minimum we have to test now second derivative test, so

f ''(x)=2/x^3

put x=1 in f ''(x)

f ''(1)=2>0 (greater than zero mean it's minimum)

f ''(-1)=-2<0 (less than zero mean it's maximum)

so x=1 is required answer .

according this you can solve easily second part.

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