求f(x)=e^(-x^2)sinx^2值域
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解决时间 2021-01-29 20:43
- 提问者网友：遁入空寂
- 2021-01-29 15:07
求f(x)=e^(-x^2)sinx^2值域
最佳答案
- 五星知识达人网友：执傲
- 2021-01-29 15:36
sinx^2 表示法有歧义， 是 sin(x^2) 还是 (sinx)^2 ？
若是 sin(x^2)， 则 f(x) = sin(x^2)/e^(x^2),
f'(x) = [2xcos(x^2)e^(x^2) - 2xsin(x^2)e^(x^2)]/[e^(x^2)]^2
= 2x[cos(x^2) - sin(x^2)]/e^(x^2)
得 驻点 x=0, x^2 = kπ+π/4, k = 0,1,2,...
f(0) = 0, f(√π/2) = 1/[√2e^(π/4)],
f[√(5π)/2] = -1/[√2e^(5π/4)],
f[√(9π)/2] = 1/[√2e^(9π/4)],
f[√(13π)/2] = -1/[√2e^(13π/4)],
f[√(17π)/2] = 1/[√2e^(17π/4)],......
得 -1/[√2e^(5π/4)] ≤ f(x) ≤ 1/[√2e^(π/4)],
若是 sin(x^2)， 则 f(x) = sin(x^2)/e^(x^2),
f'(x) = [2xcos(x^2)e^(x^2) - 2xsin(x^2)e^(x^2)]/[e^(x^2)]^2
= 2x[cos(x^2) - sin(x^2)]/e^(x^2)
得 驻点 x=0, x^2 = kπ+π/4, k = 0,1,2,...
f(0) = 0, f(√π/2) = 1/[√2e^(π/4)],
f[√(5π)/2] = -1/[√2e^(5π/4)],
f[√(9π)/2] = 1/[√2e^(9π/4)],
f[√(13π)/2] = -1/[√2e^(13π/4)],
f[√(17π)/2] = 1/[√2e^(17π/4)],......
得 -1/[√2e^(5π/4)] ≤ f(x) ≤ 1/[√2e^(π/4)],
全部回答
- 1楼网友：大漠
- 2021-01-29 16:24
虽然我很聪明，但这么说真的难到我了
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