y=(x-1)乘三次根号下x的平方,令Y的导数等于0,那么如何求X的值?

美丽谎言2007 1年前 已收到2个回答 举报
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jj_pp 幼苗

共回答了17个问题采纳率：82.4% 举报

y = (x-1)乘三次根号下x² = (x-1) * x^(2/3)y' = (x-1) * [x^(2/3)]' + (x-1)' * x^(2/3)= (x-1) * 2/3* x^(-1/3) + 1 * x^(2/3)= 2/3 (x-1) * x^(-1/3) + x^(2/3)= 2/3 (x-1) / x^(1/3) + x^(2/3)= [2/3 (x-1)...

1年前

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xin5570281 花朵

共回答了382个问题 举报

y = (x - 1) • x^(2/3)
y' = x^(2/3) + (x - 1) • (2/3)x^(-1/3)
= (5x - 2)/[3x^(1/3)]
令y' = 0
则(5x - 2)/[3x^(1/3)] = 0
=> 5x - 2 = 0
=> x = 2/5，(2/5,f(2/5))是驻点

1年前

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