Prove that d(u) is larger than 4f for u>f.
更新1:
some amendment: Prove that d(u) is larger than or equal to 4f for u>f>0.
proof
2010-10-28 7:56 am
d(u)=u^2/(u-f), where f is a constant
Prove that d(u) is larger than 4f for u>f.
Prove that d(u) is larger than 4f for u>f.
回答 (2)
2010-10-28 4:19 pm
✔ 最佳答案
d(u)=u^2/(u－f)，where f is aconstantProve that d(u) is larger than 4f for u>f.
Sol
d(u)=u^2/(u－f)，u<>f
Set d(u)=p
p=u^2/(u－f)
u^2=pu－pf
u^2－pu+pf=0
have solution
(－p)^2－4*1*pf>=0
p^2－4pf>=0
p(p－4f)>0
u<>f =>p=d(u)>0
p－4f>0
p>4f
2010-10-29 5:32 am
u^2 / (u-f) - 4f = (u^2 -4uf +4f^2)/(u-f)=(u-2f)^2/(u-f)>=0
Hence u^2/(u-f)>=4f
Hence u^2/(u-f)>=4f
收錄日期: 2021-04-23 23:24:12
原文連結 [永久失效]:
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