Evalution of a function where \(\displaystyle{t}={x}+{\frac{{{1}}}{{{x}}}}\) Consider a

London Douglas

London Douglas

Answered question

2022-04-03

Evalution of a function where t=x+1x
Consider a function
y=(x3+1x3)6(x2+1x2)+3(x+1x)
defined for real x>0. Letting t=x+1x gives:
y=t36t2+12

Answer & Explanation

pastuh7vka

pastuh7vka

Beginner2022-04-04Added 13 answers

Because by AM-GM
x+1x2x1x=2.
Your calculation of y is right:
y=t33t6(t22)+3t=t36t2+12
and you got it without using t2.

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