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  • 已知函数 , 是 的一个极值点
  • 1. 若 是 的唯一极值点,求实数 的取值范围
  • 2. 讨论 的单调性
  • 3. 若存在正数 ,使得 , 求实数 的取值范围

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  1. High Level Math
  2. GaoKao

1

已知函数 f(x)=(x−2)ex+ax2+bxf(x) = (x - 2)e^x + ax^2 + bxf(x)=(x−2)ex+ax2+bx , x=1x = 1x=1 是 f(x)f(x)f(x) 的一个极值点

1. 若 x=1x = 1x=1 是 f(x)f(x)f(x) 的唯一极值点,求实数 aaa 的取值范围

读取信息: 1. 极值点: 当 x=1x = 1x=1, f′(x)=0f^\prime(x) = 0f′(x)=0 2. 唯一极值点: 不知有何意义 3. aaa 的取值范围: 我们要求的是一个范围

尝试:

f′(x)=1⋅ex+(x−2)ex+2ax+b=ex+(x−2)ex+2ax+b=ex(1+x−2)+2ax+b=ex(x−1)+2ax+bf′(1)=0+2a+b=2a+b\begin{align*} f^\prime(x) &= 1 \cdot e^x + (x - 2)e^x + 2ax + b \\ \\ &= e^x + (x - 2)e^x + 2ax + b \\ \\ &= e^x(1 + x - 2) + 2ax + b \\ \\ &= e^x(x - 1) + 2ax + b \\ \\ f^\prime(1) &= 0 + 2a + b \\ \\ &= 2a + b \\ \\ \\ \end{align*}f′(x)f′(1)​=1⋅ex+(x−2)ex+2ax+b=ex+(x−2)ex+2ax+b=ex(1+x−2)+2ax+b=ex(x−1)+2ax+b=0+2a+b=2a+b​
f′(1)=02a+b=0Well, as long as we want a , we should keep a in the origin functionb=−2af′(x)=ex(x−1)+2ax−2aIf we want to calculate the range of a , we must make sure a is isolatedf′(x)=ex(x−1)+2a(x−1)Simplification is always good for the next step, maybe after that you’ll find a way to gof′(x)=(x−1)(ex+2a)Don’t forget, now, the x is equal to 1f′(x)=0⋅(ex+2a)\begin{align*} &f^\prime(1) = 0 \\ \\ &2a + b = 0 \\ \\ & \text{Well, as long as we want } a \text{ , we should keep } a \text{ in the origin function} \\ \\ &b = -2a \\ \\ &f^\prime(x) = e^x(x - 1) + 2ax - 2a \\ \\ & \text{If we want to calculate the range of } a \text{ , we must make sure } a \text{ is isolated} \\ \\ &f^\prime(x) = e^x(x - 1) + 2a(x - 1) \\ \\ &\text{Simplification is always good for the next step, maybe after that you'll find a way to go} \\ \\ &f^\prime(x) = (x - 1)(e^x + 2a) \\ \\ &\text{Don't forget, now, the } x \text{ is equal to } 1 \\ \\ &f^\prime(x) = 0 \cdot (e^x + 2a) \\ \\ \\ \end{align*}​f′(1)=02a+b=0Well, as long as we want a , we should keep a in the origin functionb=−2af′(x)=ex(x−1)+2ax−2aIf we want to calculate the range of a , we must make sure a is isolatedf′(x)=ex(x−1)+2a(x−1)Simplification is always good for the next step, maybe after that you’ll find a way to gof′(x)=(x−1)(ex+2a)Don’t forget, now, the x is equal to 1f′(x)=0⋅(ex+2a)​
It seems like (ex+2a) could be any value!(Just an imagination)−∞≤(ex+2a)≤+∞But we all knows, as x goes up, ex only goes +∞ , not −∞ex=+∞ , ex≠−∞And 2a can only be any value ∈(−∞,+∞) in this case(ex)+(2a)=?(+∞)+(+∞)=(+∞)(+∞)+(−∞)=(0)\begin{align*} &\text{It seems like } (e^x + 2a) \text{ could be any value!(Just an imagination)} \\ \\ &-\infty \le (e^x + 2a) \le +\infty \\ \\ & \text{But we all knows, as x goes up, } e^x \text{ only goes } +\infty \text{ , not } -\infty \\ \\ & e^x = +\infty \text{ , } e^x \not= -\infty \\ \\ & \text{And } 2a \text{ can only be any value } \in (-\infty, +\infty) \text{ in this case} \\ \\ &(e^x) + (2a) = ? \\ & (+\infty) + (+\infty) = (+\infty) \\ & (+\infty) + (-\infty) = (0) \\ \\ \\ \end{align*}​It seems like (ex+2a) could be any value!(Just an imagination)−∞≤(ex+2a)≤+∞But we all knows, as x goes up, ex only goes +∞ , not −∞ex=+∞ , ex=−∞And 2a can only be any value ∈(−∞,+∞) in this case(ex)+(2a)=?(+∞)+(+∞)=(+∞)(+∞)+(−∞)=(0)​
So (ex+2a)∈[0,+∞)(ex+2a)≥02a≥−ex2a≥−(+∞)2a≥−∞Because −∞, 0, +∞ split all number into 3 parts , Therefore 2a≥0≥−∞a≥0\begin{align*} \text{So } (e^x + 2a) &\in [0, +\infty) \\ \\ (e^x + 2a) &\ge 0 \\ \\ 2a &\ge -e^x \\ \\ 2a &\ge -(+ \infty) \\ \\ 2a &\ge -\infty \\ \\ \text{Because } -\infty &\text{, 0, } +\infty \text{ split all number into 3 parts} \text{ , Therefore } \\ \\ 2a \ge 0 \ge -\infty& \\ \\ a \ge 0& \end{align*}So (ex+2a)(ex+2a)2a2a2aBecause −∞2a≥0≥−∞a≥0​∈[0,+∞)≥0≥−ex≥−(+∞)≥−∞, 0, +∞ split all number into 3 parts , Therefore ​

2. 讨论 f(x)f(x)f(x) 的单调性

潜意识:

讨论函数的单调性,实际上就是讨论一阶导数的正负性

3. 若存在正数 x0x_0x0​ ,使得 f(x0)<af(x_0) < af(x0​)<a , 求实数 aaa 的取值范围

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