若函数f(x)=x^3-ax,若f(x)在实数集上单调递增,求实数a的取值范围
1个回答
u < v,
f(u) = u^3 - au,
f(v) = v^3 - av,
f(x)在实数集上单调递增,因此,总有,
0 > f(u) - f(v) = u^3 - au - [v^3 - av]
= u^3 - v^3 - au + av
= (u-v)(u^2 + uv + v^2) - a(u-v)
= (u-v)(u^2 + uv + v^2 - a),
而 u - v < 0,所以,总有,
u^2 + uv + v^2 - a > 0,
u^2 + uv + v^2/4 + 3v^2/4 - a > 0,
(u + v/2)^2 + 3v^2/4 - a > 0,
所以,当 a < 0时,上式总成立,满足题意.
若 a = 0,f(x) = x^3是实数集上的单调递增函数,满足题意.
若a>0.
令 u = 0,v = 0.5a^(1/2),
则,u < v,但
f(u) - f(v) = (u-v)(u^2 + uv + v^2 - a)
= （-v）(v^2 - a)
= -0.5a^(1/2)[0.25a - a]
= 0.375a^(3/2)
> 0,
与f(x)在实数集上单调递增矛盾,因此,
只有 a
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