extreme f(x)=sqrt((x-1)^2+(4-4x^2))

Lösung

extrempunkte

f (x) = (x - 1)^{2} + (4 - 4 x^{2})

Minimum (- \frac{5}{3}, 0), Maximum (- \frac{1}{3}, \frac{4}{3}), Minimum (1, 0)

Schritte zur Lösung

f^{'} (x) = - \frac{3 x + 1}{- 3 x ^{2} - 2 x + 5}

Intervalle finden:

Ansteigend

: - \frac{5}{3} < x < - \frac{1}{3},

Absteigend

: - \frac{1}{3} < x < 1

Stecke

x = - \frac{5}{3}

(x - 1)^{2} + 4 - 4 x^{2} : 0

Minimum (- \frac{5}{3}, 0)

Stecke

x = - \frac{1}{3}

(x - 1)^{2} + 4 - 4 x^{2} : \frac{4}{3}

Maximum (- \frac{1}{3}, \frac{4}{3})

Stecke

x = 1

(x - 1)^{2} + 4 - 4 x^{2} : 0

Minimum (1, 0)

Minimum (- \frac{5}{3}, 0), Maximum (- \frac{1}{3}, \frac{4}{3}), Minimum (1, 0)

e x t re m e f (x) = x^{3} + y^{3} - 30 x y

e x t re m e f (x) = \frac{5 x}{x + 1}

e x t re m e y = \frac{x ^{3} ( 3 x - 8 )}{12}

e x t re m e f (x) = e^{3 x} (7 - x)

e x t re m e \frac{( - x ^{2} - 8 x + 20 )}{( x ^{2} - 5 x ) ^{2}}