extreme f(x)= 1/2 ln((1+x)/(1-x))-2x

Lösung

extrempunkte

f (x) = \frac{1}{2} ln (\frac{1 + x}{1 - x}) - 2 x

Maximum (- \frac{1}{2}, \frac{1}{2} ln (3 - 22) + 2), Minimum (\frac{1}{2}, \frac{1}{2} ln (3 + 22) - 2)

Schritte zur Lösung

f^{'} (x) = \frac{1}{( x + 1 ) ( - x + 1 )} - 2

Intervalle finden:

Ansteigend

: - 1 < x < - \frac{1}{2},

Absteigend

: - \frac{1}{2} < x < \frac{1}{2},

Ansteigend

: \frac{1}{2} < x < 1

Stecke

x = - \frac{1}{2}

\frac{1}{2} ln (\frac{1 + x}{1 - x}) - 2 x : \frac{1}{2} ln (3 - 22) + 2

Maximum (- \frac{1}{2}, \frac{1}{2} ln (3 - 22) + 2)

Stecke

x = \frac{1}{2}

\frac{1}{2} ln (\frac{1 + x}{1 - x}) - 2 x : \frac{1}{2} ln (3 + 22) - 2

Minimum (\frac{1}{2}, \frac{1}{2} ln (3 + 22) - 2)

Maximum (- \frac{1}{2}, \frac{1}{2} ln (3 - 22) + 2), Minimum (\frac{1}{2}, \frac{1}{2} ln (3 + 22) - 2)

e x t re m e f (x) = 6 x + 12 y - x^{2} - 3 y^{2}

e x t re m e f (x) = x^{\frac{2}{5}} (x - 8)

e x t re m e f (x) = (\frac{x}{2}) + \frac{1}{x}

e x t re m e f (x) = x^{2} + 8 x + 16

e x t re m e f (x, y) = x y + \frac{4}{x} + \frac{2}{y}