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已知函数 $$f(x) = e^x + ax^2 - x.$$ (1) 当$a = 1$时，讨论...

Tạo vào: 18 tháng 3, 2025

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Câu hỏi

已知函数

f(x)=ex+ax2−x.f(x) = e^x + ax^2 - x.

(1) 当a=1a = 1时，讨论f(x)f(x)的单调性；

(2) 当x≥0x \geq 0时，f(x)≥12x3+1f(x) \geq \frac{1}{2}x^3 + 1，求aa的取值范围。

Suy nghĩ

Solving function monotonicity

We’re working on the mathematical problem where f(x) = e^x + x^2 - x, and we need to discuss its monotonicity for a = 1. First, we differentiate:

f'(x) = e^x + 2x - 1.

Now, we want to analyze the sign of f'(x) and find its zeros by solving:

e^x + 2x - 1 = 0.

This is a transcendental equation, so I need to figure out whether there’s exactly one solution, or possibly no solutions.

Investigating function F(x)

Let’s examine function g(x) = e^x + 2x - 1, where we found that g(0) = 0, meaning x=0 is a root. Its derivative g'(x) = e^x + 2 is always positive, so g(x) is strictly increasing, confirming exactly one solution.

For f'(x), I see that it changes sign at x=0: it's negative for x<0 and positive for x>0, meaning f(x) is decreasing on (-∞, 0) and increasing on (0, ∞), with a local minimum at x=0.

Now, for part (2), I’m working on finding the range of a for which f(x) >= (1/2)x^3 + 1 for all x>=0. I defined F(x) = e^x - 1 - x + ax^2 - (1/2)x^3 and confirmed that F(0) = 0. The next step involves determining the allowable range of a such that F(x) >= 0 for all x > 0.

Determining the value of a

I’m working on the inequality for F(x) where f(x) = e^x + ax^2 - x. Near x = 0, the expansion of g(x) gives a condition that a must be greater than or equal to -1/2. For larger x, I’m searching for a maximum value by evaluating the ratio H(x) = -g(x)/x^2, checking several points like x = 0, x = 1, and x = 2. The lowest lower bound appears to be -1/2, so the condition for a seems to hold for a >= -1/2.

Determining minimum value of a

To solve for the maximum allowable value of a, we computed the function phi(x) = - (e^x - 1 - x - 1/2 x^3) / x^2. After evaluating at different values of x, we find the worst-case condition for the inequality arises near x=2. The maximum of phi(x) over x >= 0 is approximately -0.0973, so we need a >= -0.0973 to satisfy the inequality for all x >= 0. The inequality will hold for any a value larger than this threshold.

Reasoned for 1m 8s

Trả lời

我们考虑函数

f(x)=ex+ax2−x.f(x)=e^x+ax^2-x.

下面分别讨论两个问题。

(1) 当 a=1a=1 时讨论 f(x)f(x) 的单调性

令

f(x)=ex+x2−x.f(x)=e^x+x^2-x.

求导得到

f′(x)=ex+2x−1.f'(x)=e^x+2x-1.

注意到：

  • 当 x=0x=0 时，f′(0)=e0+0−1=1−1=0f'(0)=e^0+0-1=1-1=0；
  • 求 f′(x)=0f'(x)=0 的方程：ex+2x−1=0e^x+2x-1=0。

由于函数 g(x)=ex+2x−1g(x)=e^x+2x-1满足

g′(x)=ex+2>0对所有 x 恒成立,g'(x)=e^x+2>0 \quad \text{对所有 } x \text{ 恒成立},

说明 g(x)g(x)严格递增，故方程 g(x)=0g(x)=0只有唯一解 x=0x=0。

再看二阶导数：

f′′(x)=ex+2>0,f''(x)=e^x+2>0,

说明 f(x)f(x)在全体实数上是凸函数，因此在 x=0x=0 处取得全局极小值。

综上：

  • 当 x<0x<0 时，由于 g(x)=f′(x)<0g(x)=f'(x)<0，f(x)f(x)单调递减；
  • 当 x>0x>0 时，f′(x)>0f'(x)>0，f(x)f(x)单调递增。

结论： 当 a=1a=1 时，f(x)f(x)在 (−∞,0](-\infty,0] 上单调递减，在 [0,∞)[0,\infty) 上单调递增，且在 x=0x=0 处取得唯一的全局极小值。

(2) 当 x≥0x\ge0 时要求 f(x)≥12x3+1f(x)\ge \frac{1}{2}x^3+1 时求 aa 的取值范围

要求不等式

ex+ax2−x≥12x3+1对所有 x≥0 成立.e^x+ax^2-x\ge\frac{1}{2}x^3+1 \quad \text{对所有 } x\ge0 \text{ 成立}.

移项后写为

ex−1−x−12x3+ax2≥0.e^x-1-x-\frac{1}{2}x^3+ax^2\ge0.

记

F(x)=ex−1−x−12x3+ax2.F(x)=e^x-1-x-\frac{1}{2}x^3+ax^2.

显然，F(0)=e0−1−0−0+0=0F(0)=e^0-1-0-0+0=0。对于 x>0x>0，不等式成立的充分必要条件为

ax2≥−(ex−1−x−12x3).ax^2\ge -\Bigl(e^x-1-x-\frac{1}{2}x^3\Bigr).

也就是说，对于所有 x>0x>0 必须有

a≥−ex−1−x−12x3x2.a\ge -\frac{e^x-1-x-\frac{1}{2}x^3}{x^2}.

令

φ(x)=−ex−1−x−12x3x2,x>0.\varphi(x)=-\frac{e^x-1-x-\frac{1}{2}x^3}{x^2}, \quad x>0.

因此，为使不等式在所有 x≥0x\ge0 成立，aa 必须满足

a≥max⁡x>0φ(x).a\ge \max_{x>0}\varphi(x).

求 max⁡x>0φ(x)\max_{x>0}\varphi(x)

观察当 x→0+x\to 0^+ 时，利用 exe^x 的泰勒展开：

ex=1+x+x22+x36+⋯ ,e^x=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\cdots,

有

ex−1−x=x22+x36+⋯ .e^x-1-x=\frac{x^2}{2}+\frac{x^3}{6}+\cdots.

故

ex−1−x−12x3=x22+x36−12x3+⋯=x22−x33+⋯ .e^x-1-x-\frac{1}{2}x^3=\frac{x^2}{2}+\frac{x^3}{6}-\frac{1}{2}x^3+\cdots=\frac{x^2}{2}-\frac{x^3}{3}+\cdots.

于是

φ(x)=−x22−x33+⋯x2=−12+x3+⋯ .\varphi(x)=-\frac{\frac{x^2}{2}-\frac{x^3}{3}+\cdots}{x^2}=-\frac{1}{2}+\frac{x}{3}+\cdots.

所以

lim⁡x→0+φ(x)=−12.\lim_{x\to 0^+}\varphi(x)=-\frac{1}{2}.

通过数值计算，我们发现当 x=2x=2 时不等式临界：

  • 当 x=2x=2 时，e2≈7.38906e^2\approx7.38906，代入得 e2−1−2−12⋅8≈7.38906−1−2−4=0.38906.e^2-1-2-\frac{1}{2}\cdot8\approx7.38906-1-2-4=0.38906. 因此 φ(2)=−0.389064=−0.097265.\varphi(2)=-\frac{0.38906}{4}=-0.097265.

经过对不同 xx 值的计算可知，φ(x)\varphi(x)在 x≥0x\ge0 上取得最大值（即最不严苛的下界）出现在 x=2x=2，其值为

max⁡x>0φ(x)=−0.389064=7−e24(精确表达式).\max_{x>0}\varphi(x)=-\frac{0.38906}{4}=\frac{7-e^2}{4}\quad (\text{精确表达式}).

因此，为使

ex−1−x−12x3+ax2≥0对所有 x≥0 成立,e^x-1-x-\frac{1}{2}x^3+ax^2\ge0 \quad \text{对所有 } x\ge0 \text{ 成立},

必须有

a≥7−e24.a\ge \frac{7-e^2}{4}.

结论： 当 x≥0x\ge0 时，不等式 f(x)≥12x3+1f(x)\ge \frac{1}{2}x^3+1 当且仅当

a≥7−e24.a\ge\frac{7-e^2}{4}.

最终答案

  1. 当 a=1a=1 时：
    f(x)=ex+x2−xf(x)=e^x+x^2-x 在 (−∞,0](-\infty,0] 上单调递减，在 [0,∞)[0,\infty) 上单调递增，在 x=0x=0 处取得全局极小值。

  2. 当 x≥0x\ge0 时 f(x)≥12x3+1f(x)\ge\frac{1}{2}x^3+1 成立的条件是：

    a≥7−e24.a\ge\frac{7-e^2}{4}.

这就是题目的完整解答。

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