數學問題 幫忙謝謝?
回答 (2)
2017-08-05 2:14 am
a^2+2b
=(2b-1)^2 + 2b
=4b^2 - 4b + 1 + 2b
=4b^2 - 2b + 1
, coefficient of b^2 is 4, greater than 0 ===> only have minimum value
===> Δ = (-2)^2 - 4*4*1 = -12
Optimum value 極值= -Δ/(4a) = -(-12)/(4*4) = 0.75
Minimum value is 0.75
=(2b-1)^2 + 2b
=4b^2 - 4b + 1 + 2b
=4b^2 - 2b + 1
, coefficient of b^2 is 4, greater than 0 ===> only have minimum value
===> Δ = (-2)^2 - 4*4*1 = -12
Optimum value 極值= -Δ/(4a) = -(-12)/(4*4) = 0.75
Minimum value is 0.75
2017-08-03 10:58 pm
a^2+b^2
=(2b-1)^2+b^2
=5b^2-4b+1
=5(b-2/5)^2+1-4/5
=5(b-2/5)^2+1/5
當 a=-1/5, b=2/5 時, 有極小值 1/5
=(2b-1)^2+b^2
=5b^2-4b+1
=5(b-2/5)^2+1-4/5
=5(b-2/5)^2+1/5
當 a=-1/5, b=2/5 時, 有極小值 1/5
收錄日期: 2021-05-01 01:11:25
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20170803141850AAXC3Yu