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# AS level Maths Question HELP! watch

1. Needed help with this question,

Prove that the minimum possible value of the sum of a positive real number and it's reciprocal is 2

Help is much appreciated
2. Let r be a positive real number. 1/r is its reciprocal. The sum

s =r + 1/r = (r^2+1)/r

Assume s<2, then r^2+1<2r ==? r^2-2r+1<0

then (r-1)^2<0, NO! (r-1)^2 ge 0, so we have a contradiciton, hence the assertion

r + 1/r >= 2 holds.
3. Thanks but I'm not quite sure how you got that answer, could you tell me how you got r + 1/r = (r²+1)/r ?
4. (Original post by Medic900)
Let r be a positive real number. 1/r is its reciprocal. The sum

s =r + 1/r = (r^2+1)/r

Assume s<2, then r^2+1<2r ==? r^2-2r+1<0

then (r-1)^2<0, NO! (r-1)^2 ge 0, so we have a contradiciton, hence the assertion

r + 1/r >= 2 holds.
Thanks but how did you got the first part, r + 1/r = (r²+1)/r?
5. We have f(x)=x+1/x, just set f'(x)=0 and find min point and find the value of f at this point.
6. Let f(X)= X+ (1/X) where X>0 (as given)1st derivative of f(X) with respect to X => 1+ (-1/x^2) Let us now find the turning point So 1+ (-1/x^2) = 0And we get X= 1 or -1Since X>0 , we can only take X=1But X= 1 can be Maxima or minima So we do the 2nd derivative test which gives us that X=1 is minima .Therefore minimum value => f(1) = 2
7. (Original post by Dingan123)
Let f(X)= X+ (1/X) where X>0 (as given)1st derivative of f(X) with respect to X => 1+ (-1/x^2) Let us now find the turning point So 1+ (-1/x^2) = 0And we get X= 1 or -1Since X>0 , we can only take X=1But X= 1 can be Maxima or minima So we do the 2nd derivative test which gives us that X=1 is minima .Therefore minimum value => f(1) = 2
Thank you for that

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