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Let \(f\left( x \right) = {e^{ - {x^2}}}\)
1. Find the horizontal asymptote of the graph of \(f\)
2. Find the inflection point of\(f\)
3. Sketch the graph of\(f\)
【師範大學102年度轉學考】 |
詳解：\(\lim \limits_{x \to \infty } f\left( x \right) = \lim \limits_{x \to \infty } {e^{ - {x^2}}} = 0\)表示有一條水平漸近線\(y = 0\)
\(f\left( x \right) = {e^{ - {x^2}}} \Rightarrow f'\left( x \right) = ( - 2x){e^{ - {x^2}}} \Rightarrow f''\left( x \right) = ( - 2){e^{ - {x^2}}} + (4{x^2}){e^{ - {x^2}}} = (4{x^2} - 2){e^{ - {x^2}}}\)
二階臨界值\(x = \frac{1}{{\sqrt 2 }}\)，\( x = \frac{{ - 1}}{{\sqrt 2 }}\)
| \(x\) |
|
\(-\frac{1}{\sqrt 2 }\) |
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\(\frac{1}{\sqrt 2 }\) |
|
| \(f''(x)\) |
+ |
\(\vdots \) |
- |
\(\vdots \) |
+ |
|
| \(f(x)\) |
\(\cup\) |
\(\vdots \) |
\(\cap\) |
\(\vdots \) |
\(\cup\) |
有兩個反曲點\((\frac{1}{{\sqrt 2 }} , \frac{1}{{\sqrt e }})\)，\(( - \frac{1}{{\sqrt 2 }} , \frac{1}{{\sqrt e }})\)
概圖
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