Предмет: Алгебра, автор: vasiat

Из точки А(0;-1) проведите касательную к графику функции y= ln (1/3 e^3 x )

Ответы

Автор ответа: mefody66

$y=ln(1/3*e^{3x}) = ln(1/3) + ln(e^{3x}) = -ln(3) +3x$
Это прямая, и касательная к ней - эта же самая прямая.
Но она не проходит через точку A(0; -1).
Так что задача решения не имеет.

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