extreme (2x^2-5x+5)/(x-2)

Lösung

extrempunkte

\frac{2 x ^{2} - 5 x + 5}{x - 2}

Maximum (\frac{4 - 6}{2}, - \frac{4 3 - 3 2}{2}), Minimum (\frac{4 + 6}{2}, \frac{4 3 + 3 2}{2})

Schritte zur Lösung

f^{'} (x) = \frac{2 x ^{2} - 8 x + 5}{( x - 2 ) ^{2}}

Intervalle finden:

Ansteigend

: - \infty < x < \frac{4 - 6}{2},

Absteigend

: \frac{4 - 6}{2} < x < 2,

Absteigend

: 2 < x < \frac{4 + 6}{2},

Ansteigend

: \frac{4 + 6}{2} < x < \infty

Stecke

x = \frac{4 - 6}{2}

\frac{2 x ^{2} - 5 x + 5}{x - 2} : - \frac{4 3 - 3 2}{2}

Maximum (\frac{4 - 6}{2}, - \frac{4 3 - 3 2}{2})

Stecke

x = \frac{4 + 6}{2}

\frac{2 x ^{2} - 5 x + 5}{x - 2} : \frac{4 3 + 3 2}{2}

Minimum (\frac{4 + 6}{2}, \frac{4 3 + 3 2}{2})

Maximum (\frac{4 - 6}{2}, - \frac{4 3 - 3 2}{2}), Minimum (\frac{4 + 6}{2}, \frac{4 3 + 3 2}{2})

e x t re m e f (x) = 2 x^{2} - 12 x + 20

e x t re m e f (x) = 4 sin^{2} (x), 0 \leq x \leq π

e x t re m e f (x, y) = x^{3} + y^{3} - 3 x^{2} - 9 y^{2} - 9 x

e x t re m e f (x, y) = 4 x^{2} + \frac{1}{4} y^{3} - 2 x y

e x t re m e f (x) = 8 x + 7 x^{- 1}