extreme f(x)=(x^2+2)/(x+1)

Lösung

extrempunkte

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

Maximum (- 1 - 3, - \frac{6 + 2 3}{3}), Minimum (- 1 + 3, \frac{6 - 2 3}{3})

Schritte zur Lösung

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

Intervalle finden:

Ansteigend

: - \infty < x < - 1 - 3,

Absteigend

: - 1 - 3 < x < - 1,

Absteigend

: - 1 < x < - 1 + 3,

Ansteigend

: - 1 + 3 < x < \infty

Stecke

x = - 1 - 3

\frac{x ^{2} + 2}{x + 1} : - \frac{6 + 2 3}{3}

Maximum (- 1 - 3, - \frac{6 + 2 3}{3})

Stecke

x = - 1 + 3

\frac{x ^{2} + 2}{x + 1} : \frac{6 - 2 3}{3}

Minimum (- 1 + 3, \frac{6 - 2 3}{3})

Maximum (- 1 - 3, - \frac{6 + 2 3}{3}), Minimum (- 1 + 3, \frac{6 - 2 3}{3})

e x t re m e f (x) = 2 x - 11 ln (8 x), 2 \leq x \leq 9

e x t re m e f (x) = 3 x^{\frac{2}{3}}, - 1 \leq x \leq 1

e x t re m e (2 x - x^{2}) (2 y - y^{2})

e x t re m e sin (2 x)

e x t re m e f (x, y) = \frac{1}{3} x^{3} - 5 x^{2} - y^{2} - 3 y