已知函数
f ( x ) = ( x − 2 ) e x + a x 2 + b x f(x) = (x - 2)e^x + ax^2 + bx f ( x ) = ( x − 2 ) e x + a x 2 + b x ，
x = 1 x = 1 x = 1 是
f ( x ) f(x) f ( x ) 的一个极值点
1. 若
x = 1 x = 1 x = 1 是
f ( x ) f(x) f ( x ) 的唯一极值点，求实数
a a a 的取值范围
读取信息： 1. 极值点： 当 x = 1 x = 1 x = 1 f ′ ( x ) = 0 f^\prime(x) = 0 f ′ ( x ) = 0 a a a
尝试：
f ′ ( x ) = 1 ⋅ e x + ( x − 2 ) e x + 2 a x + b = e x + ( x − 2 ) e x + 2 a x + b = e x ( 1 + x − 2 ) + 2 a x + b = e x ( x − 1 ) + 2 a x + b f ′ ( 1 ) = 0 + 2 a + b = 2 a + b \begin{align*}
f^\prime(x) &= 1 \cdot e^x + (x - 2)e^x + 2ax + b
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&= e^x + (x - 2)e^x + 2ax + b
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&= e^x(1 + x - 2) + 2ax + b
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&= e^x(x - 1) + 2ax + b
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f^\prime(1) &= 0 + 2a + b
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&= 2a + b
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\end{align*} f ′ ( x ) f ′ ( 1 ) = 1 ⋅ e x + ( x − 2 ) e x + 2 a x + b = e x + ( x − 2 ) e x + 2 a x + b = e x ( 1 + x − 2 ) + 2 a x + b = e x ( x − 1 ) + 2 a x + b = 0 + 2 a + b = 2 a + b f ′ ( 1 ) = 0 2 a + b = 0 Well, as long as we want a , we should keep a in the origin function b = − 2 a f ′ ( x ) = e x ( x − 1 ) + 2 a x − 2 a If we want to calculate the range of a , we must make sure a is isolated f ′ ( x ) = e x ( x − 1 ) + 2 a ( x − 1 ) Simplification is always good for the next step, maybe after that you’ll find a way to go f ′ ( x ) = ( x − 1 ) ( e x + 2 a ) Don’t forget, now, the x is equal to 1 f ′ ( x ) = 0 ⋅ ( e x + 2 a ) \begin{align*}
&f^\prime(1) = 0
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&2a + b = 0
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& \text{Well, as long as we want } a \text{ , we should keep } a \text{ in the origin function}
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&b = -2a
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&f^\prime(x) = e^x(x - 1) + 2ax - 2a
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& \text{If we want to calculate the range of } a \text{ , we must make sure } a \text{ is isolated}
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&f^\prime(x) = e^x(x - 1) + 2a(x - 1)
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&\text{Simplification is always good for the next step, maybe after that you'll find a way to go}
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&f^\prime(x) = (x - 1)(e^x + 2a)
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&\text{Don't forget, now, the } x \text{ is equal to } 1
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&f^\prime(x) = 0 \cdot (e^x + 2a)
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\end{align*} f ′ ( 1 ) = 0 2 a + b = 0 Well, as long as we want a , we should keep a in the origin function b = − 2 a f ′ ( x ) = e x ( x − 1 ) + 2 a x − 2 a If we want to calculate the range of a , we must make sure a is isolated f ′ ( x ) = e x ( x − 1 ) + 2 a ( x − 1 ) Simplification is always good for the next step, maybe after that you’ll find a way to go f ′ ( x ) = ( x − 1 ) ( e x + 2 a ) Don’t forget, now, the x is equal to 1 f ′ ( x ) = 0 ⋅ ( e x + 2 a ) It seems like ( e x + 2 a ) could be any value!(Just an imagination) − ∞ ≤ ( e x + 2 a ) ≤ + ∞ But we all knows, as x goes up, e x only goes + ∞ , not − ∞ e x = + ∞ , e x ≠ − ∞ And 2 a can only be any value ∈ ( − ∞ , + ∞ ) in this case ( e x ) + ( 2 a ) = ? ( + ∞ ) + ( + ∞ ) = ( + ∞ ) ( + ∞ ) + ( − ∞ ) = ( 0 ) \begin{align*}
&\text{It seems like } (e^x + 2a) \text{ could be any value!(Just an imagination)}
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&-\infty \le (e^x + 2a) \le +\infty
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& \text{But we all knows, as x goes up, } e^x \text{ only goes } +\infty \text{ , not } -\infty
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& e^x = +\infty \text{ , } e^x \not= -\infty
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& \text{And } 2a \text{ can only be any value } \in (-\infty, +\infty) \text{ in this case}
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&(e^x) + (2a) = ?
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& (+\infty) + (+\infty) = (+\infty)
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& (+\infty) + (-\infty) = (0)
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\end{align*} It seems like ( e x + 2 a ) could be any value!(Just an imagination) − ∞ ≤ ( e x + 2 a ) ≤ + ∞ But we all knows, as x goes up, e x only goes + ∞ , not − ∞ e x = + ∞ , e x = − ∞ And 2 a can only be any value ∈ ( − ∞ , + ∞ ) in this case ( e x ) + ( 2 a ) = ? ( + ∞ ) + ( + ∞ ) = ( + ∞ ) ( + ∞ ) + ( − ∞ ) = ( 0 ) So ( e x + 2 a ) ∈ [ 0 , + ∞ ) ( e x + 2 a ) ≥ 0 2 a ≥ − e x 2 a ≥ − ( + ∞ ) 2 a ≥ − ∞ Because − ∞ , 0, + ∞ split all number into 3 parts , Therefore 2 a ≥ 0 ≥ − ∞ a ≥ 0 \begin{align*}
\text{So } (e^x + 2a) &\in [0, +\infty)
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(e^x + 2a) &\ge 0
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2a &\ge -e^x
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2a &\ge -(+ \infty)
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2a &\ge -\infty
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\text{Because } -\infty &\text{, 0, } +\infty \text{ split all number into 3 parts} \text{ , Therefore }
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2a \ge 0 \ge -\infty&
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a \ge 0&
\end{align*} So ( e x + 2 a ) ( e x + 2 a ) 2 a 2 a 2 a Because − ∞ 2 a ≥ 0 ≥ − ∞ a ≥ 0 ∈ [ 0 , + ∞ ) ≥ 0 ≥ − e x ≥ − ( + ∞ ) ≥ − ∞ , 0, + ∞ split all number into 3 parts , Therefore 潜意识:
Copy 讨论函数的单调性，实际上就是讨论一阶导数的正负性