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

Lösung

extrempunkte

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

Maximum (5 - 2, \frac{5 2 - 4}{2}), Minimum (5 + 2, \frac{4 + 5 2}{2})

Schritte zur Lösung

f^{'} (x) = \frac{x ^{2} - 10 x + 23}{( x - 5 ) ^{2}}

Intervalle finden:

Ansteigend

: - \infty < x < 5 - 2,

Absteigend

: 5 - 2 < x < 5,

Absteigend

: 5 < x < 5 + 2,

Ansteigend

: 5 + 2 < x < \infty

Stecke

x = 5 - 2

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

Maximum (5 - 2, \frac{5 2 - 4}{2})

Stecke

x = 5 + 2

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

Minimum (5 + 2, \frac{4 + 5 2}{2})

Maximum (5 - 2, \frac{5 2 - 4}{2}), Minimum (5 + 2, \frac{4 + 5 2}{2})

e x t re m e (8 - x) (x + 1)^{2}

e x t re m e f (x) = x^{3} + 11 x + 10

e x t re m e f (x) = 3 sin (x), 0 \leq x \leq 2 π

e x t re m e \frac{e ^{x}}{8 + e ^{x}}

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