函数f(x)=√ax*2-1 在点x=2处的导数值为4/√7 则a=?
人气:199 ℃ 时间:2020-03-28 17:34:11
解答
f(x) = √a x² - 1
f'(x) = 2√a x
f'(2) = 4/√7
2√a (2) = 4/√7
√a = √(1/7)
a = 1/7忘记说了根号是包括全部数字的f(x) = √(ax^2 - 1)f'(x) = 1/[2√(ax^2 - 1)] * (a * 2x)= ax/√(ax^2 - 1)f'(2) = 4/√7a(2)/√(a(2)^2 - 1) = 4/√72a/√(4a - 1) = 4/√72a√7/4 = √(4a - 1)28a^2/16 = 4a - 128a^2 = 64a - 167a^2 - 16a + 4 = 0(7a - 2)(a - 2) = 0a = 2 或 a = 2/7
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