extreme f(x)=(16x-20)*ln(4x-5)

Lösung

extrempunkte

f (x) = (16 x - 20) \cdot ln (4 x - 5)

Minimum (\frac{1 + 5 e}{4 e}, \frac{4}{e} ln (\frac{1 + 5 e}{e} - 5))

Schritte zur Lösung

f^{'} (x) = 16 (ln (4 x - 5) + 1)

Intervalle finden:

Absteigend

: \frac{5}{4} < x < \frac{1 + 5 e}{4 e},

Ansteigend

: \frac{1 + 5 e}{4 e} < x < \infty

Stecke

x = \frac{1 + 5 e}{4 e}

(16 x - 20) \cdot ln (4 x - 5) : \frac{4}{e} ln (\frac{1 + 5 e}{e} - 5)

Minimum (\frac{1 + 5 e}{4 e}, \frac{4}{e} ln (\frac{1 + 5 e}{e} - 5))

e x t re m e f (x) = 3 x^{4} - 6 x^{3} + 9

e x t re m e 2300 x - 2 x^{2}

e x t re m e f (x) = x^{2} + x y + y^{2} - 34 y + 385

e x t re m e f (x) = x^{3} - \frac{3}{2} x^{2}, - 5 \leq x \leq 2

e x t re m e f (x) = lo g_{10} (x^{2} + 1)