extreme f(x)=x^2-7x+10

Lösung

extrempunkte

f (x) = x^{2} - 7 x + 10

Minimum (\frac{7}{2}, - \frac{9}{4})

Schritte zur Lösung

f^{'} (x) = 2 x - 7

Intervalle finden:

Absteigend

: - \infty < x < \frac{7}{2},

Ansteigend

: \frac{7}{2} < x < \infty

Stecke

x = \frac{7}{2}

x^{2} - 7 x + 10 : - \frac{9}{4}

Minimum (\frac{7}{2}, - \frac{9}{4})

e x t re m e f (x) = 1 - 9 x - 6 x^{2} - x^{3}

e x t re m e y = \frac{4 x ^{2}}{x ^{2} - 4}

e x t re m e f (x, y) = x^{3} + 3 x y^{2} - 3 y^{2} + 3 x^{2} + 11

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

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