| 4 | ||
y= | e2x−1 | |
| 5 |
| 5 | e2x | 5ye | 5ye | ||||
y = | ⇔ e2x = | ⇔ 2x = ln( | ) ⇔ | ||||
| 4 | e | 4 | 4 |
| 1 | 5ye | |||
⇔ x = | ln( | ) | ||
| 2 | 4 |
| 1 | 5x | |||
f−1(x) = | (ln | + 1) | ||
| 2 | 4 |
| 5y | ||
2x − 1 = ln | ||
| 4 |
| 5y | ||
2x= 1+ ln | ||
| 4 |
| 5y | ||
ostatecznie x = 0,5 + 0,5*ln | . ⇔ y = 0,5 + ln √5x/4 funkcja odwrotna | |
| 4 |
| √5x | 1 | |||
No to moze jeszcze tak: f−1(x) = ln( | ) + | |||
| 2 | 2 |